[LeetCode]Insert Interval

最新推荐文章于 2017-01-10 06:40:16 发布

CiaoLiang

最新推荐文章于 2017-01-10 06:40:16 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/CiaoLiang/article/details/49994647

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本文讨论了如何在已排序的区间集合中插入一个新区间，并在必要时进行合并。通过实例展示了插入过程及合并规则。

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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].

[思路]

这道题目思路还是很简单的，把间隔插入到合适位置，然后再进行有效合并。

实现中要注意各种CornerCase。比如第一个位置插入，最后插入，没有合并的插入。

/**
 * 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) {
         int first = newInterval.start;
         int second = newInterval.end;
         bool flag = false;
         vector<Interval> res;
         if(intervals.size()==0){
             res.push_back(newInterval);
             return res;//一边是空集
         }
         for(int i=0; i<intervals.size(); ++i){
             Interval temp;
             if(i!=0) temp = res.back();
             if(i!=0&&!(intervals[i].start > temp.end || intervals[i].end < temp.start)){ //和前面重复的合并
                 temp.end = max(temp.end,intervals[i].end);
                 temp.start = min(temp.start,intervals[i].start);
                 res.pop_back();
                 res.push_back(temp);
                 continue;
             }
             
             if(!(intervals[i].start > second || intervals[i].end < first)){
                 //和插入序列如果能合并
                 intervals[i].end = max(intervals[i].end ,second);
                 intervals[i].start = min(intervals[i].start ,first);
                 flag = true;//判断是否插入能合并
             }
             
             res.push_back(intervals[i]);
         }
         if(flag == false){//把newInterval插入到合适位置
             if(newInterval.start>res.back().start){
                 res.push_back(newInterval);
                 return res;
             }
             for(int i=0; i<res.size(); ++i){
                 if(newInterval.start<res[i].start){
                     res.insert(res.begin()+i,newInterval);
                     break;
                 }
             }
         }
         
         return res;
    }
};