73 lines
2.1 KiB
Java
73 lines
2.1 KiB
Java
package com.leetcode.array;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.Comparator;
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import java.util.List;
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/**
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*
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* 给出一个区间的集合，请合并所有重叠的区间。
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示例 1:
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输入: intervals = [[1,3],[2,6],[8,10],[15,18]]
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输出: [[1,6],[8,10],[15,18]]
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解释: 区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].
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示例 2:
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输入: intervals = [[1,4],[4,5]]
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输出: [[1,5]]
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解释: 区间 [1,4] 和 [4,5] 可被视为重叠区间。
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注意：输入类型已于2019年4月15日更改。 请重置默认代码定义以获取新方法签名。
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提示：
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intervals[i][0] <= intervals[i][1]
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作者：力扣 (LeetCode)
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链接：https://leetcode-cn.com/leetbook/read/array-and-string/c5tv3/
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来源：力扣（LeetCode）
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著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
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*
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*/
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public class ArrayMerge {
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public int[][] merge(int[][] intervals) {
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if (intervals.length == 0) {
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return new int[0][2];
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}
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Arrays.sort(intervals, new Comparator<int[]>() {
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public int compare(int[] interval1, int[] interval2) {
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return interval1[0] - interval2[0];
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}
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});
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List<int[]> merged = new ArrayList<int[]>();
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for (int i = 0; i < intervals.length; ++i) {
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int L = intervals[i][0], R = intervals[i][1];
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if (merged.size() == 0 || merged.get(merged.size() - 1)[1] < L) {
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merged.add(new int[]{L, R});
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} else {
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merged.get(merged.size() - 1)[1] = Math.max(merged.get(merged.size() - 1)[1], R);
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}
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}
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return merged.toArray(new int[merged.size()][]);
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}
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public static void main(String[] args) {
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ArrayMerge merge = new ArrayMerge();
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int[][] intervals = {
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{1, 4},
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{1, 4}
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};
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int[][] res = merge.merge(intervals);
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for (int i = 0; i < res.length; i++) {
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System.out.println("[" + res[i][0] + "," + res[i][1] + "]");
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}
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}
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}
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