双指针

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortedSquares = function(nums) {
   var result = [];
    //遍历A中每个数字
   for(var i = 0; i < nums.length; i ++){
        result.push(nums[i] * nums[i]);
    }
    //对result的元素排序
   result.sort((a,b) => a - b);
   return result;
};

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortedSquares = function(nums) {
    let res = []
    for (let i = 0, j = nums.length - 1; i <= j;) {
      const left = Math.abs(nums[i])
      const right = Math.abs(nums[j])
      if (right > left) {
        // push element to the front of the array
        res.unshift(right * right)

        j--
      } else {
        res.unshift(left * left)

        i++
      }
    }
    return res

};

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {void} Do not return anything, modify nums in-place instead.
 */
 const reverse = (nums, start, end) => {
    while (start < end) {
        const temp = nums[start];
        nums[start] = nums[end];
        nums[end] = temp;
        start += 1;
        end -= 1;
    }
}
var rotate = function(nums, k) {
    k %= nums.length;
    reverse(nums, 0, nums.length - 1);
    reverse(nums, 0, k - 1);
    reverse(nums, k, nums.length - 1);

};

/**
 * @param {number[]} nums
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var moveZeroes = function(nums) { 
let j=0;
 for(var i=0;i<nums.length;i++){
     if(nums[i]!=0){        
        nums[i] = nums.splice(j, 1, nums[i])[0]
        j=j+1
     }

 }
};

/**
 * @param {number[]} numbers
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(numbers, target) {
    let left=0,right=numbers.length-1
    var arr=[]
    while(left<right){
      if(numbers[left]+numbers[right]<target){
           left++
       }
      else if(numbers[left]+numbers[right]>target){
           right--
       }
      else{
        arr[0]=left+1
        arr[1]= right+1
        return arr
      }
   }

};

我的解答

/**
 * @param {string} s
 * @return {string}
 */
  const reverse = (nums, start, end) => {
    while (start < end) {
        const temp = nums[start];
        nums[start] = nums[end];
        nums[end] = temp;
        start += 1;
        end -= 1;

    }
    return nums
}
var reverseWords = function(s) {
 let left=0,right=0
 while(right<s.length){
 if(s[right]==' '){
   s=reverse(s,left,right-1)
   left=right+1
 }
 if(right==s.length-1){
   s=reverse(s,left,right)
 }
 right++
 console.log(left,right)
 }
 return s
};

然而输出：

原因：Js中的String类型不可变！！！
新的解答（新建一个字符串）

/**
 * @param {string} s
 * @return {string}
 */
var reverseWords = function(s) {
 let left=0,right=0
 let arr=[]
 while(right<s.length){
 if(s[right]==' '){
    for(var i=right-1;i>=left;i--){
        arr.push(s[i])
    }
    arr.push(' ') 
   left=right+1
 }
 if(right==s.length-1){
    for(var i=right;i>=left;i--){
        arr.push(s[i])
    }

 }
 right++
 }
 return arr.join('')
};

解法1

* @param {ListNode} head
 * @return {ListNode}
 */
var middleNode = function(head) {
    slow = fast = head;
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }
    return slow;
};

解法2

var middleNode = function(head) {
    let A = [head];
    while (A[A.length - 1].next != null)
        A.push(A[A.length - 1].next);
    return A[Math.trunc(A.length / 2)];
};

 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function(head, n) {
    let slow = head
    let fast = head

     while(n--!=0){
            fast=fast.next;
        }
        // 如果快指针走到了最后说明删除的是第一个节点,就返回head.next就好
        if(fast==null){
            return head.next;
        }
        // 使得slow每次都是在待删除的前一个节点, 所以要先让fast先走一步
        fast=fast.next;
        while(fast!=null){
            fast=fast.next;
            slow=slow.next;
        }
        // 因为已经保证了是待删除节点的前一个节点, 直接删除即可
        slow.next=slow.next.next;
        return head;

};