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57. Insert Interval
Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).
You may assume that the intervals were initially sorted according to their start times.
Example 1:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5] Output: [[1,5],[6,9]]
Example 2:
Input: intervals =[[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval =[4,8]Output: [[1,2],[3,10],[12,16]] Explanation: Because the new interval[4,8]overlaps with[3,5],[6,7],[8,10].
NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
解题分析
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2019-10-23 12:27
*/
public int[][] insert(int[][] intervals, int[] newInterval) {
int left = newInterval[0];
int right = newInterval[1];
boolean flag = false;
List<int[]> result = new ArrayList<>();
for (int[] ints : intervals) {
if (right < ints[0]) {
// 在插入区间的右侧且无交集
if (!flag) {
// 注意：这里是 left, right，不能直接放 newInterval
result.add(new int[]{left, right});
flag = true;
}
result.add(ints);
} else if (ints[1] < left) {
// 在插入区间的左侧且无交集
result.add(ints);
} else {
left = Math.min(left, ints[0]);
right = Math.max(right, ints[1]);
}
}
if (!flag) {
result.add(new int[]{left, right});
}
return result.toArray(new int[result.size()][2]);
}