Remove Covered Intervals (M)
題目
Given a list of intervals, remove all intervals that are covered by another interval in the list.
Interval [a,b) is covered by interval [c,d) if and only if c <= a and b <= d.
After doing so, return the number of remaining intervals.
Example 1:
Input: intervals = [[1,4],[3,6],[2,8]]
Output: 2
Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.
Example 2:
Input: intervals = [[1,4],[2,3]]
Output: 1
Example 3:
Input: intervals = [[0,10],[5,12]]
Output: 2
Example 4:
Input: intervals = [[3,10],[4,10],[5,11]]
Output: 2
Example 5:
Input: intervals = [[1,2],[1,4],[3,4]]
Output: 1
Constraints:
1 <= intervals.length <= 1000intervals[i].length == 20 <= intervals[i][0] < intervals[i][1] <= 10^5- All the intervals are unique.
題意
洗掉陣列中所有被其他區間覆寫的區間，
思路
將陣列按照左端點從小到大排序(相同則右端點大的在前)，從左到右遍歷，記錄右端點的最大值rightMax，如果當前區間的右端點小于rightMax，說明該區間被之前的某個區間覆寫，將其刪去；否則更新rightMax，
代碼實作
Java
class Solution {
public int removeCoveredIntervals(int[][] intervals) {
int cnt = 0;
Arrays.sort(intervals, (int[] a, int[] b) -> a[0] < b[0] ? -1 : a[0] == b[0] ? b[1] - a[1] : 1);
int rightMax = intervals[0][1];
for (int i = 1; i < intervals.length; i++) {
if (intervals[i][1] <= rightMax) {
cnt++;
} else {
rightMax = intervals[i][1];
}
}
return intervals.length - cnt;
}
}
轉載請註明出處，本文鏈接：https://www.uj5u.com/qita/156020.html
標籤：其他