252. Meeting Rooms

包含判断两个区间相交的基本操作：

  1. var canAttendMeetings = function(intervals) {
  2.     let result = true;
  4.     intervals.sort((a, b) => a[0] - b[0]); // 必须先排序
  5.     for (let i = 1; i < intervals.length; i++) {
  6.         if (didOverlap(intervals[i - 1], intervals[i])) {
  7.             result = false;
  8.             break;
  9.         }
  10.     }
  12.     function didOverlap(range1, range2) { // 判断两个区间是否重叠
  13.         let [left1, right1] = range1;
  14.         let [left2, right2] = range2;
  16.         if (right1 <= left2 || right2 <= left1) { // 取条件好找的
  17.             return false;
  18.         }
  19.         return true;
  20.     }
  22.     return result;
  23. };

判断区间重叠的方式可以更抽象：
image.png

  1.     function didOverlap(range1, range2) { // 判断两个区间是否重叠
  2.         let [left1, right1] = range1;
  3.         let [left2, right2] = range2;
  5.         return Math.max(left1, left2) < Math.min(right1, right2);
  6.     }

57. Insert Interval

题解：https://leetcode-cn.com/problems/insert-interval/solution/shou-hua-tu-jie-57-cha-ru-qu-jian-fen-cheng-3ge-ji/

  1. var insert = function(intervals, newInterval) {
  2.     let n = intervals.length;
  3.     if (n === 0) {
  4.         return [newInterval];
  5.     }
  7.     let [left, right] = newInterval;
  9.     let ans = [];
  10.     let i = 0;
  12.     // 第一段
  13.     while (i < n) {
  14.         let [_left, _right] = intervals[i];
  15.         if (_right < left) {
  16.             ans.push(intervals[i]);
  17.             i++;
  18.         } else {
  19.             break;
  20.         }
  21.     }
  23.     // 第二段
  24.     let newLeft = left;
  25.     let newRight = right;
  26.     while(i < n) {
  27.         let [_left, _right] = intervals[i];
  28.         if (_left <= right) { // 相交，这个条件判断很牛啊
  29.             newLeft = Math.min(_left, left, newLeft);
  30.             newRight = Math.max(_right, right, newRight);
  31.             i++;
  32.         } else {
  33.             break;
  34.         }
  35.     }
  36.     ans.push([newLeft, newRight]);
  38.     // 第三段
  39.     while(i < n) { // 后面直接收割
  40.         ans.push(intervals[i]);
  41.         i++;
  42.     }
  44.     return ans;
  45. };

1272. Remove Interval

删除区间，利用求交集

  1. var removeInterval = function(intervals, toBeRemoved) {
  2.     let [left, right] = toBeRemoved;
  3.     if (left >= right) {
  4.         return intervals; // 啥也不干
  5.     }
  6.     let ans = [];
  8.     for (let interval of intervals) {
  9.         let [_left, _right] = interval;
  10.         if (right <= _left || _right <= left) { // 不相交
  11.             ans.push(interval);
  12.         } else { // 这里的写法很有特色，重点记忆一下
  13.             if (_left < left) {
  14.                 ans.push([_left, left])
  15.             }
  16.             if (right < _right) {
  17.                 ans.push([right, _right]);
  18.             }
  19.         }
  20.     }
  22.     return ans;
  23. };