56. Merge Intervals
56. Merge Intervals
题目
Given an array of intervals
where intervals[i] = [starti, endi]
, merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.
Example 1:
Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].
Example 2:
Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
Constraints:
1 <= intervals.length <= 10^4
intervals[i].length == 2
0 <= starti <= endi <= 10^4
题目大意
以数组 intervals
表示若干个区间的集合,其中单个区间为 intervals[i] = [starti, endi]
。请你合并所有重叠的区间,并返回 一个不重叠的区间数组,该数组需恰好覆盖输入中的所有区间 。
解题思路
- 一个区间可以表示为
[start, end]
,先按区间的start
排序 - 对于几个相交区间合并后的结果区间
x
,x.start
一定是这些相交区间中start
最小的,x.end
一定是这些相交区间中end
最大的 - 由于已经排了序,
x.start
很好确定,求 x.end
也很容易,可以类比在数组中找最大值的过程。
代码
/**
* @param {number[][]} intervals
* @return {number[][]}
*/
var merge = function (intervals) {
let res = [];
// 先按区间的 `start` 排序
intervals = intervals.sort((a, b) => a[0] - b[0]);
for (let i = 0; i < intervals.length; i++) {
let start = intervals[i][0],
end = intervals[i][1];
// 如果区间相交,求结果区间 [start, end]
while (intervals[i + 1] && inRange(intervals[i + 1], start, end)) {
start = Math.min(start, intervals[i + 1][0]);
end = Math.max(end, intervals[i + 1][1]);
i++;
}
res.push([start, end]);
}
return res;
};
// 判断区间是否相交
var inRange = function (interval, start, end) {
let [m, n] = interval;
if (m < start && n < start) return false;
if (m > end && n > end) return false;
return true;
};
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