LeetCode——Insert Interval

最新推荐文章于 2025-12-06 07:43:37 发布

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本文介绍了一种算法，用于在一个已排序的非重叠区间集合中插入一个新的区间，并在必要时进行合并。通过两个实例展示了如何将新区间[2,5]和[4,9]分别插入到给定的区间集合中，并返回合并后的结果。

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].

» Solve this problem

我偷懒了，直接用了Merge Intervals的代码。稍作修改即可。

实际上，这题可以用二分来做，不过处理起来有点烦。

而且最坏的时间复杂度依旧是O(n)。

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
     vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<Interval> answer;
        intervals.push_back(newInterval);
        int min = 1000000, max = -1000000;
        for (vector<Interval>::iterator iter = intervals.begin(); iter != intervals.end(); iter++) {
            min = min < iter->start ? min : iter->start;
            max = max > iter->end ? max : iter->end;
        }
        
        int left[max - min + 1];
        int right[max - min + 1];
        memset(left, 0, sizeof(int) * (max - min + 1));
        memset(right, 0, sizeof(int) * (max - min + 1));
        for (vector<Interval>::iterator iter = intervals.begin(); iter != intervals.end(); iter++) {
            left[iter->start - min]++;
            right[iter->end - min]++;
        }
        
        int start = -1, count = 0;
        for (int i = 0; i < max - min + 1; i++) {
            count += left[i];
            if (count > 0 && start == -1) {
                start = i;
            }
            count -= right[i];
            if (count == 0 && start != -1) {
                answer.push_back(Interval(start + min, i + min));
                start = -1;
            }
        }
        
        return answer;
    }    
};