【057-Insert Interval(插入区间)】【LeetCode-面试算法经典-Java实现】【所有题目目录索引】原题
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: Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9]. Example 2: Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16]. This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
题目大意
给定一系列非覆盖的区间,插入一个新的区间,有必要的时候进行区间合并,,区间开始是以起始时间进行合并的
解题思路
如果原来的区间比插入区间小就插入结果集,如果插入区间有重叠,更新插入区间,如果插入区间小于原来的区间,先插入插入区间,再添加大的区间
代码实现
算法实现类
import java.util.LinkedList;import java.util.List;{public List<Interval> insert(List<Interval> intervals, Interval newInterval) {// 保存结果的集合List<Interval> result = new LinkedList<>();// 输入集非空if (intervals != null) {// 遍历元素for (Interval item : intervals) {(newInterval == null || item.end < newInterval.start) {result.add(item);}(item.start > newInterval.end) {result.add(newInterval);result.add(item);newInterval = null;}// 插入区间有重叠,更新插入区间else {newInterval.start = Math.min(newInterval.start, item.start);newInterval.end = Math.max(newInterval.end, item.end);}}}// 如果插入区间非空说明插入区间还未被处理if (newInterval != null) {result.add(newInterval);}return result;}}评测结果
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