双指针法

具体列子

1. 返回倒数第K个节点

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解法：

  1. /**
  2.  * 1. 暴力法：遍历链表，存到字符串中，读取倒数第K个节点
  3.  * 2. 快慢指针法
  4.  */
  5. var kthToLast = function(head, k) {
  6.   let fastNode = head
  7.   for (let i = 0; i < k; i++) {
  9.     fastNode = fastNode.next
  10.   }
  11.   let ans = head
  12.   while (fastNode) {
  13.     fastNode = fastNode.next
  14.     ans = ans.next
  15.   }
  16.   return ans.val
  17. };

2. 验证回文字符串（LeetCode-125）

image.png
解法：

  1. /**
  2.  * @param {string} s
  3.  * @return {boolean}
  4.  */
  5. var isPalindrome = function(s) {
  6.   const normalizeS = s.replace(/\W/g, '').toLowerCase()
  7.   let start = 0
  8.   let end = normalizeS.length - 1
  9.   while (start < end) {
  10.     if (normalizeS[start] !== normalizeS[end]) return false
  11.     start++
  12.     end--
  13.   }
  14.   return true
  15. };

3. 环形链表（LeetCode-141）

image.png
解法：

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} head
 * @return {boolean}
 */
var hasCycle = function(head) {
  if (!head) return false
  let fast = head.next
  let slow = head
  while(fast && slow) {
    if (fast === slow) return true
    slow = slow.next
    fast = fast.next ? fast.next.next : null
  }
  return false
};

4. 实现 strStr()（LeetCode-28）

image.png
解法：

/**
 * @param {string} haystack
 * @param {string} needle
 * @return {number}
 */
var strStr = function(haystack, needle) {
  if (haystack === needle) return 0
  const needleLen = needle.length
  let start = 0
  let end = needleLen
  const len = haystack.length
  if (len < needleLen) return -1
  while(end <= len) {
    if (haystack.slice(start, end) === needle) return start
    start++
    end++
  }
  return -1
};

5. 反转字符串中的元音字母（LeetCode-345）

image.png
解法：

/**
 * @param {string} s
 * @return {string}
 */
var reverseVowels = function (s) {
    s = s.split("");
    let u = new Set("aeiouAEIOU".split(""));
    let l = 0, r = s.length - 1;
    while (true) {
        while (l < r && l < s.length && r >= 0 && !u.has(s[l])) l++;
        while (l < r && l < s.length && r >= 0 && !u.has(s[r])) r--;
        if (l >= r) {
            break;
        }
        let t = s[l];
        s[l] = s[r];
        s[r] = t;
        l++;
        r--;
    }
    return s.join("");
};

6. 移动零（LeetCode-283）

image.png
解法：

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

7. 合并两个有序数组（LeetCode-88）

解法：

/**
 * @param {number[]} nums1
 * @param {number} m
 * @param {number[]} nums2
 * @param {number} n
 * @return {void} Do not return anything, modify nums1 in-place instead.
 */
var merge = function(nums1, m, nums2, n) {
  if (!n) return
  let count = m + n - 1
  let l1 = m > 1 ? m - 1 : 0
  let l2 = n > 1 ? n - 1 : 0
  while (l1 >= 0 && l2 >= 0) {
    nums1[count--] = nums1[l1] < nums2[l2] ? nums2[l2--] : nums1[l1--]
  }
  if (l2 >= 0) {
    for (let i = 0; i <= l2; i++) {
      nums1[i] = nums2[i]
    }
  }
};