0435. Non-overlapping Intervals (M)
Non-overlapping Intervals (M)
题目
Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
Example 1:
Input: [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.
Example 2:
Input: [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of intervals non-overlapping.
Example 3:
Input: [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
Note:
- You may assume the interval's end point is always bigger than its start point.
- Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.
题意
给定一组区间,要求删去最少的区间,使剩余的区间互不重叠。
思路
贪心,类似于规划区间使不重叠的区间最多。先将区间按照左端点升序排序,再依次遍历,如果有两两重叠,则删去右端点较大的那个区间。
代码实现
Java
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
Arrays.sort(intervals, (int[] a, int[] b) -> a[0] - b[0]);
int count = 0;
int i = 0;
for (int j = 1; j < intervals.length; j++) {
if (intervals[i][1] > intervals[j][0]) {
count++;
i = intervals[i][1] >= intervals[j][1] ? j : i;
} else {
i = j;
}
}
return count;
}
}
JavaScript
/**
* @param {number[][]} intervals
* @return {number}
*/
var eraseOverlapIntervals = function (intervals) {
if (intervals.length === 0) {
return 0
}
let addCount = 0
intervals.sort((a, b) => a[1] - b[1])
let i = 0
addCount++
for (let j = 1; j < intervals.length; j++) {
if (intervals[j][0] >= intervals[i][1]) {
addCount++
i = j
}
}
return intervals.length - addCount
}