【LeetCode】Insert Interval 解题报告

最新推荐文章于 2023-01-02 19:08:26 发布

ljiabin

最新推荐文章于 2023-01-02 19:08:26 发布

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分类专栏：算法研究 LeetCode解题报告文章标签： Interval BinarySearch 二分查找 LeetCode

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这篇博客主要讲解如何在已按开始时间排序且不重叠的LeetCode区间问题中，使用二分查找法插入新区间并处理可能的重叠，通过示例详细解释了插入和合并的逻辑。

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【题目】

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].

【解析】

题意：给一些已经按开始时间排好序的区间，现在往里面插入一个区间，如果有重叠就合并区间，返回插入后的区间序列。

思路：先插入，在合并重叠区间。既然是已经排好序的，就可以用二分查找的方法，把要插入的这个区间放到应该的位置。合并重叠区间的方法《【LeetCode】Merge Intervals 解题报告》一样。

【Java代码】

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
        List<Interval> ans = new ArrayList<Interval>();
        
        // insert newInterval by binary searching
        int l = 0;
        int r = intervals.size() - 1;
        while (l <= r) {
            int mid = (l + r) >> 1;
            if (intervals.get(mid).start > newInterval.start) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        intervals.add(l, newInterval);
        
        // merge all overlapping intervals
        int start = intervals.get(0).start;
        int end = intervals.get(0).end;
        for (int i = 1; i < intervals.size(); i++) {
            Interval inter = intervals.get(i);
            if (inter.start > end) {
                ans.add(new Interval(start, end));
                start = inter.start;
                end = inter.end;
            } else {
                end = Math.max(end, inter.end);
            }
        }
        ans.add(new Interval(start, end));
        
        return ans;
    }
}