更新0435 无重叠区间 java版本右边界解法中排序的判断逻辑

This commit is contained in:
zhenghao
2022-04-20 10:23:18 +08:00
parent 0cfd92f587
commit f9146a3b98

View File

@ -184,13 +184,14 @@ public:
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> {
if (a[0] == a[0]) return a[1] - b[1];
return a[0] - b[0];
// 按照区间右边界升序排序
return a[1] - b[1];
});
int count = 0;
int edge = Integer.MIN_VALUE;
for (int i = 0; i < intervals.length; i++) {
// 若上一个区间的右边界小于当前区间的左边界,说明无交集
if (edge <= intervals[i][0]) {
edge = intervals[i][1];
} else {