搞定大厂算法面试之leetcode精讲14.排序算法

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常见排序算法复杂度

d s

n^2除nlogn在不同数据规模下的结果

ds_114

常见排序算法

算法可视化来源:http://visualgo.net/

冒泡排序:时间复杂度O(n^2)

  • 比较相邻元素,如果第一个比第二个大,则交换他们
  • 一轮下来,可以保证最后一个数是最大的
  • 执行n-1轮,就可以完成排序

ds_192

function bubbleSort(arr) {
    var len = arr.length;
    for (var i = 0; i < len; i++) {
        for (var j = 0; j < len - 1 - i; j++) {
            if (arr[j] > arr[j+1]) {        //相邻元素两两对比
                var temp = arr[j+1];        //元素交换
                arr[j+1] = arr[j];
                arr[j] = temp;
            }
        }
    }
    return arr;
}

选择排序:时间复杂度O(n^2)

  • 找到数组中的最小值,将它放在第一位
  • 接着找到第二小的值,将它放在第二位
  • 依次类推,执行n-1轮

ds_193

function selectionSort(arr) {
    var len = arr.length;
    var minIndex, temp;
    for (var i = 0; i < len - 1; i++) {
        minIndex = i;
        for (var j = i + 1; j < len; j++) {
            if (arr[j] < arr[minIndex]) {     //寻找最小的数
                minIndex = j;                 //将最小数的索引保存
            }
        }
        temp = arr[i];
        arr[i] = arr[minIndex];
        arr[minIndex] = temp;
    }
    return arr;
}

插入排序:时间复杂度O(n^2)

  • 从第二个数开始往前比
  • 比它大就往后排
  • 以此类推直到最后一个数

ds_194

function insertionSort(arr) {
    var len = arr.length;
    var preIndex, current;
    for (var i = 1; i < len; i++) {
        preIndex = i - 1;
        current = arr[i];
        while(preIndex >= 0 && arr[preIndex] > current) {
            arr[preIndex+1] = arr[preIndex];
            preIndex--;
        }
        arr[preIndex+1] = current;
    }
    return arr;
}

归并排序:时间复杂度O(nlogn),分的时间复杂度O(logn),合并的过程的复杂度是O(n)

  • 分:把数组分成两半,递归子数组,进行分割操作,直到分成一个数

  • 合:把两个字数组合并成一个有序数组,直到全部子数组合并完毕,合并前先准备一个空数组,存放合并之后的结果,然后不断取出两个子数组的第一个元素,比较他们的大小,小的先进入之前准备的空数组中,然后继续遍历其他元素,直到子数组中的元素都完成遍历

ds_195

function mergeSort(arr) {  //采用自上而下的递归方法
    var len = arr.length;
    if(len < 2) {
        return arr;
    }
    var middle = Math.floor(len / 2),
        left = arr.slice(0, middle),
        right = arr.slice(middle);
    return merge(mergeSort(left), mergeSort(right));
}

function merge(left, right)
{
    var result = [];

    while (left.length && right.length) {
        if (left[0] <= right[0]) {
            result.push(left.shift());
        } else {
            result.push(right.shift());
        }
    }

    while (left.length)
        result.push(left.shift());

    while (right.length)
        result.push(right.shift());

    return result;
}

快速排序:时间复杂度O(nlogn),递归复杂度是O(logn),分区复杂度O(n)

  • 分区:从数组中选一个基准值,比基准值小的放在它的前面,比基准值大的放在它的后面
  • 递归:递归对基准值前后的子数组进行第一步的操作

ds_196

function quickSort(arr, left, right) {
    var len = arr.length,
        partitionIndex,
        left = typeof left != 'number' ? 0 : left,
        right = typeof right != 'number' ? len - 1 : right;

    if (left < right) {
        partitionIndex = partition(arr, left, right);
        quickSort(arr, left, partitionIndex-1);
        quickSort(arr, partitionIndex+1, right);
    }
    return arr;
}

function partition(arr, left ,right) {     //分区操作
  	//设定基准值位置(pivot)当然也可以选择最右边的元素为基准 也可以随机选择然后和最左或最右元素交换
    var pivot = left,                      
        index = pivot + 1;
    for (var i = index; i <= right; i++) {
        if (arr[i] < arr[pivot]) {
            swap(arr, i, index);
            index++;
        }        
    }
    swap(arr, pivot, index - 1);
    return index-1;
}

function swap(arr, i, j) {
    var temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}

215. 数组中的第K个最大元素 (medium)

方法1.维护大小为k的小顶堆,当堆的元素个数小于等于k时,遍历数组,让数组的元素不断加入堆,当堆的大小大于k时,让堆顶元素出列,遍历完数组之后,小顶堆堆顶的元素就是第k大元素。

复杂度:时间复杂度O(nlogk),循环n次,每次堆的操作是O(logk)。空间复杂度O(k)

ds_209

js:

class Heap {
    constructor(comparator = (a, b) => a - b, data = []) {
        this.data = data;
        this.comparator = comparator;//比较器
        this.heapify();//堆化
    }

    heapify() {
        if (this.size() < 2) return;
        for (let i = Math.floor(this.size() / 2) - 1; i >= 0; i--) {
            this.bubbleDown(i);//bubbleDown操作
        }
    }

    peek() {
        if (this.size() === 0) return null;
        return this.data[0];//查看堆顶
    }

    offer(value) {
        this.data.push(value);//加入数组
        this.bubbleUp(this.size() - 1);//调整加入的元素在小顶堆中的位置
    }

    poll() {
        if (this.size() === 0) {
            return null;
        }
        const result = this.data[0];
        const last = this.data.pop();
        if (this.size() !== 0) {
            this.data[0] = last;//交换第一个元素和最后一个元素
            this.bubbleDown(0);//bubbleDown操作
        }
        return result;
    }

    bubbleUp(index) {
        while (index > 0) {
            const parentIndex = (index - 1) >> 1;//父节点的位置
            //如果当前元素比父节点的元素小,就交换当前节点和父节点的位置
            if (this.comparator(this.data[index], this.data[parentIndex]) < 0) {
                this.swap(index, parentIndex);//交换自己和父节点的位置
                index = parentIndex;//不断向上取父节点进行比较
            } else {
                break;//如果当前元素比父节点的元素大,不需要处理
            }
        }
    }

    bubbleDown(index) {
        const lastIndex = this.size() - 1;//最后一个节点的位置
        while (true) {
            const leftIndex = index * 2 + 1;//左节点的位置
            const rightIndex = index * 2 + 2;//右节点的位置
            let findIndex = index;//bubbleDown节点的位置
            //找出左右节点中value小的节点
            if (
                leftIndex <= lastIndex &&
                this.comparator(this.data[leftIndex], this.data[findIndex]) < 0
            ) {
                findIndex = leftIndex;
            }
            if (
                rightIndex <= lastIndex &&
                this.comparator(this.data[rightIndex], this.data[findIndex]) < 0
            ) {
                findIndex = rightIndex;
            }
            if (index !== findIndex) {
                this.swap(index, findIndex);//交换当前元素和左右节点中value小的
                index = findIndex;
            } else {
                break;
            }
        }
    }

    swap(index1, index2) {//交换堆中两个元素的位置
        [this.data[index1], this.data[index2]] = [this.data[index2], this.data[index1]];
    }

    size() {
        return this.data.length;
    }
}



var findKthLargest = function (nums, k) {
    const h = new Heap((a, b) => a - b);

    for (const num of nums) {
        h.offer(num);//加入堆
        if (h.size() > k) {//堆的size超过k时,出堆
            h.poll();
        }
    }

    return h.peek();
};

方法2:堆排序

ds_156

  • 思路:利用原地堆排序的思想,将前k-1大的元素加入队尾,最后队顶的元素就是第k大的元素
  • 复杂度:时间复杂度O(nlogn),堆的创建复杂度是O(n),移动前k-1大的元素然后堆化复杂度是O(klogn),k<=n,最差的情况下是O(nlogn),空间复杂度O(logn),递归的栈空间

js:

var findKthLargest = function (nums, k) {
    let heapSize = nums.length;
    buildMaxHeap(nums, heapSize); //构建大顶堆 大小为heapSize
    //大顶堆 前k-1个堆顶元素不断和数组的末尾元素交换 然后重新heapify堆顶元素
    //这个操作就是之前小顶堆出堆顶的操作,只不过现在是原地排序
    for (let i = nums.length - 1; i >= nums.length - k + 1; i--) {
        swap(nums, 0, i);//交换堆顶和数组末尾元素
        --heapSize; //堆大小减1
        maxHeapify(nums, 0, heapSize);//重新heapify
    }
    return nums[0];//返回堆顶元素,就是第k大的元素

    function buildMaxHeap(nums, heapSize) {
        for (let i = Math.floor(heapSize / 2) - 1; i >= 0; i--) {//从第一个非叶子节点开始构建
            maxHeapify(nums, i, heapSize);
        }
    }
    // 从左向右,自上而下的调整节点
    function maxHeapify(nums, i, heapSize) {
        let l = i * 2 + 1;//左节点
        let r = i * 2 + 2;//右节点
        let largest = i;
        if (l < heapSize && nums[l] > nums[largest]) {
            largest = l;
        }
        if (r < heapSize && nums[r] > nums[largest]) {
            largest = r;
        }
        if (largest !== i) {
            swap(nums, i, largest); //找到左右节点中大的元素交换
            //递归交换后面的节点
            maxHeapify(nums, largest, heapSize);
        }
    }

    function swap(a, i, j) {//交换函数
        let temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }
};

java:

class Solution {
    public int findKthLargest(int[] nums, int k) {
        int heapSize = nums.length;
        buildMaxHeap(nums, heapSize);
        for (int i = nums.length - 1; i >= nums.length - k + 1; --i) {
            swap(nums, 0, i);
            --heapSize;
            maxHeapify(nums, 0, heapSize);
        }
        return nums[0];
    }

    public void buildMaxHeap(int[] a, int heapSize) {
        for (int i = heapSize / 2; i >= 0; --i) {
            maxHeapify(a, i, heapSize);
        } 
    }

    public void maxHeapify(int[] a, int i, int heapSize) {
        int l = i * 2 + 1, r = i * 2 + 2, largest = i;
        if (l < heapSize && a[l] > a[largest]) {
            largest = l;
        } 
        if (r < heapSize && a[r] > a[largest]) {
            largest = r;
        }
        if (largest != i) {
            swap(a, i, largest);
            maxHeapify(a, largest, heapSize);
        }
    }

    public void swap(int[] a, int i, int j) {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }
}

方法3:快速排序的分区方法

ds_157

  • 思路:借鉴快排的思路,不断随机选择基准元素,看进行partition之后,该元素是不是在n-k的位置。

  • 复杂度:

    1. 时间复杂度O(nlogn)
    2. 空间复杂度O(logn),递归的深度

js:

const findKthLargest = (nums, k) => {
    const n = nums.length;

    const quick = (l, r) => {
        if (l > r) return;//递归终止条件
        let random = Math.floor(Math.random() * (r - l + 1)) + l; //随机选取一个索引
        swap(nums, random, r); //将它和位置r的元素交换,让nums[r]作为基准元素

        //对基准元素进行partition
        let pivotIndex = partition(nums, l, r);
        //进行partition之后,基准元素左边的元素都小于它 右边的元素都大于它
        //如果partition之后,这个基准元素的位置pivotIndex正好是n-k 则找大了第k大的数
        //如果n-k<pivotIndex,则在pivotIndex的左边递归查找
        //如果n-k>pivotIndex,则在pivotIndex的右边递归查找
        if (n - k < pivotIndex) {
            quick(l, pivotIndex - 1);
        } else {
            quick(pivotIndex + 1, r);
        }
    };

    quick(0, n - 1);//函数开始传入的left=0,right= n - 1
    return nums[n - k]; //最后找到了正确的位置 也就是n-k等于pivotIndex 这个位置的元素就是第k大的数
};

function partition(nums, left, right) {
    let pivot = nums[right];             	//最右边的元素为基准
    let pivotIndex = left;               	//pivotIndex初始化为left
    for (let i = left; i < right; i++) { 	//遍历left到right-1的元素
        if (nums[i] < pivot) {             	//如果当前元素比基准元素小
            swap(nums, i, pivotIndex);       	//把它交换到pivotIndex的位置
            pivotIndex++;                    	//pivotIndex往前移动一步
        }
    }
    swap(nums, right, pivotIndex);       	//最后交换pivotIndex和right
    return pivotIndex;                   	//返回pivotIndex
}

function swap(nums, p, q) {//交换数组中的两个元素
    const temp = nums[p];
    nums[p] = nums[q];
    nums[q] = temp;
}


java:

class Solution {
    Random random = new Random();

    public int findKthLargest(int[] nums, int k) {
        return quickSelect(nums, 0, nums.length - 1, nums.length - k);
    }

    public int quickSelect(int[] a, int l, int r, int index) {
        int q = randomPartition(a, l, r);
        if (q == index) {
            return a[q];
        } else {
            return q < index ? quickSelect(a, q + 1, r, index) : quickSelect(a, l, q - 1, index);
        }
    }

    public int randomPartition(int[] a, int l, int r) {
        int i = random.nextInt(r - l + 1) + l;
        swap(a, i, r);
        return partition(a, l, r);
    }

    public int partition(int[] a, int l, int r) {
        int x = a[r], i = l - 1;
        for (int j = l; j < r; ++j) {
            if (a[j] <= x) {
                swap(a, ++i, j);
            }
        }
        swap(a, i + 1, r);
        return i + 1;
    }

    public void swap(int[] a, int i, int j) {
        int temp = a[i];
        a[i] = a[j];
        a[j] = temp;
    }
}

148. 排序链表(medium)

ds_199

动画过大,点击查看

方法1:自顶向下
  • 思路:用归并排序的思路,先不断分割,知道每个子区间只有一个节点位置,然后开始合并。
  • 复杂度:时间复杂度O(nlogn),和归并排序的复杂度一样。空间复杂度O(logn),递归的栈空间

js:

const merge = (head1, head2) => {
    const dummyHead = new ListNode(0);
    let temp = dummyHead, temp1 = head1, temp2 = head2;
    while (temp1 !== null && temp2 !== null) {//合并子区间 小的节点先连
        if (temp1.val <= temp2.val) {
            temp.next = temp1;
            temp1 = temp1.next;
        } else {
            temp.next = temp2;
            temp2 = temp2.next;
        }
        temp = temp.next;
    }
    if (temp1 !== null) {//两条链表还有节点没合并完,直接合并过来
        temp.next = temp1;
    } else if (temp2 !== null) {
        temp.next = temp2;
    }
    return dummyHead.next;
}

const toSortList = (head, tail) => {
    if (head === null) {//极端情况
        return head;
    }
    if (head.next === tail) {//分割到只剩一个节点
        head.next = null;
        return head;
    }
    let slow = head, fast = head;
    while (fast !== tail) {//的到中间节点
        slow = slow.next;
        fast = fast.next;
        if (fast !== tail) {
            fast = fast.next;
        }
    }
    const mid = slow;
    return merge(toSortList(head, mid), toSortList(mid, tail));//分割区间 递归合并
}

var sortList = function(head) {
    return toSortList(head, null);
};


java:

class Solution {
    public ListNode sortList(ListNode head) {
        return toSortList(head, null);
    }

    public ListNode toSortList(ListNode head, ListNode tail) {
        if (head == null) {
            return head;
        }
        if (head.next == tail) {
            head.next = null;
            return head;
        }
        ListNode slow = head, fast = head;
        while (fast != tail) {
            slow = slow.next;
            fast = fast.next;
            if (fast != tail) {
                fast = fast.next;
            }
        }
        ListNode mid = slow;
        ListNode list1 = toSortList(head, mid);
        ListNode list2 = toSortList(mid, tail);
        ListNode sorted = merge(list1, list2);
        return sorted;
    }

    public ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummyHead = new ListNode(0);
        ListNode temp = dummyHead, temp1 = head1, temp2 = head2;
        while (temp1 != null && temp2 != null) {
            if (temp1.val <= temp2.val) {
                temp.next = temp1;
                temp1 = temp1.next;
            } else {
                temp.next = temp2;
                temp2 = temp2.next;
            }
            temp = temp.next;
        }
        if (temp1 != null) {
            temp.next = temp1;
        } else if (temp2 != null) {
            temp.next = temp2;
        }
        return dummyHead.next;
    }
}

方法2:自底向上

  • 思路:直接进行循环合并操作。
  • 复杂度:时间复杂度O(nlogn),空间复杂度O(1)

js:

const merge = (head1, head2) => {
    const dummyHead = new ListNode(0);
    let temp = dummyHead, temp1 = head1, temp2 = head2;
    while (temp1 !== null && temp2 !== null) {
        if (temp1.val <= temp2.val) {
            temp.next = temp1;
            temp1 = temp1.next;
        } else {
            temp.next = temp2;
            temp2 = temp2.next;
        }
        temp = temp.next;
    }
    if (temp1 !== null) {
        temp.next = temp1;
    } else if (temp2 !== null) {
        temp.next = temp2;
    }
    return dummyHead.next;
}

var sortList = function(head) {
    if (head === null) {
        return head;
    }
    let length = 0;
    let node = head;
    while (node !== null) {
        length++;
        node = node.next;
    }
    const dummyHead = new ListNode(0, head);
    for (let subLength = 1; subLength < length; subLength <<= 1) {
        let prev = dummyHead, curr = dummyHead.next;
        while (curr !== null) {
            let head1 = curr;
            for (let i = 1; i < subLength && curr.next !== null; i++) {
                curr = curr.next;
            }
            let head2 = curr.next;
            curr.next = null;
            curr = head2;
            for (let i = 1; i < subLength && curr != null && curr.next !== null; i++) {
                curr = curr.next;
            }
            let next = null;
            if (curr !== null) {
                next = curr.next;
                curr.next = null;
            }
            const merged = merge(head1, head2);
            prev.next = merged;
            while (prev.next !== null) {
                prev = prev.next;
            }
            curr = next;
        }
    }
    return dummyHead.next;
};

java:

class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null) {
            return head;
        }
        int length = 0;
        ListNode node = head;
        while (node != null) {
            length++;
            node = node.next;
        }
        ListNode dummyHead = new ListNode(0, head);
        for (int subLength = 1; subLength < length; subLength <<= 1) {
            ListNode prev = dummyHead, curr = dummyHead.next;
            while (curr != null) {
                ListNode head1 = curr;
                for (int i = 1; i < subLength && curr.next != null; i++) {
                    curr = curr.next;
                }
                ListNode head2 = curr.next;
                curr.next = null;
                curr = head2;
                for (int i = 1; i < subLength && curr != null && curr.next != null; i++) {
                    curr = curr.next;
                }
                ListNode next = null;
                if (curr != null) {
                    next = curr.next;
                    curr.next = null;
                }
                ListNode merged = merge(head1, head2);
                prev.next = merged;
                while (prev.next != null) {
                    prev = prev.next;
                }
                curr = next;
            }
        }
        return dummyHead.next;
    }

    public ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummyHead = new ListNode(0);
        ListNode temp = dummyHead, temp1 = head1, temp2 = head2;
        while (temp1 != null && temp2 != null) {
            if (temp1.val <= temp2.val) {
                temp.next = temp1;
                temp1 = temp1.next;
            } else {
                temp.next = temp2;
                temp2 = temp2.next;
            }
            temp = temp.next;
        }
        if (temp1 != null) {
            temp.next = temp1;
        } else if (temp2 != null) {
            temp.next = temp2;
        }
        return dummyHead.next;
    }
}
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