leetcode刷题之路——20190807（160、168）

[yangxy6]

160

/*
 * @lc app=leetcode id=160 lang=javascript
 *
 * [160] Intersection of Two Linked Lists
 */
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 * 时间复杂度O(n)，空间复杂度O(1)
 * 思路：想法特别精妙，两个链表长度不一致，短+长拼接和长+短拼接，最后节点相等时不是相交节点就是（null===null）
 * 举栗子：[0,9,1,2,4]和[3,2,4] 两个链表拼接后[0,9,1,2,4,3,2,4]和[3,2,4,0,9,1,2,4]相交节点在2开始相同
 */
var getIntersectionNode = function(headA, headB) {
  if (headA === null || headB === null) return null;
  let curA = headA;
  let curB = headB;
  // curA===curB说明相交或者全部节点走完直接返回
  while (curA !== curB) {
    //A节点走完走B 实现A+B
    curA = curA === null ? headB : curA.next;
    curB = curB === null ? headA : curB.next;
  }
  return curA;
};

168

/*
 * @lc app=leetcode id=168 lang=javascript
 * √ 18/18 cases passed (48 ms)
 * √ Your runtime beats 87.86 % of javascript submissions
 * √ Your memory usage beats 14.29 % of javascript submissions (34 MB)
 * [168] Excel Sheet Column Title
 */
/**
 * @param {number} n
 * @return {string}
 * 思路：循环n/26直到n为0，count % 26 === 0正好是26倍数时需要count/26-1，同时将Z放到map第一位。
 */
var convertToTitle = function(n) {
  const map = ['Z', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y'];
  let ans = '';
  let count = n;

  while (count) {
    ans = map[count % 26] + ans;
    count = count % 26 === 0 ? Math.floor(count / 26) - 1 : Math.floor(count / 26);
  }
  return ans;
};
console.log(convertToTitle(701));