Wednesday, November 26, 2014

[LeetCode] 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:
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:

题目涉及两个问题:
1. 如何判断两个interval有重合。
2. 如果重合,如何合并。
3. 如何判断[s1, e1]是否在[s2, e2]前面

A1. 两个interval有重合有很多种可能性,但判断它们不重合则比较简单直观。无非两种情况:

(1) [s1, e1]  [s2, e2]:即s2>e1 
(2) [s2, e2]  [s1, e1]:即s1>e2

不重合的条件为:s2>e1 || s1>e2,反之则重合。

A2. 对于两个有重合的interval: [s1, e1],[s2, e2]。合并后变为[min(s1,s2), max(e1,e2)]。

A3. [s1, e1]在[s2, e2]前面的条件:e1<s2

所以插入interval a的算法为:扫描队列中每个interval I[i]:
如果a已经被插入了,则只要插入I(i)就行。
如果a在I(i)前,则先插入a再插入I(i)到结果。
如果a和I(i)有重合,则将I(i)合并入a,但并不插入结果。
如果a在I(i)后,则插入I(i)到结果。


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class Solution {
public:
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        vector<Interval> ret;
        bool isInsert = false;
        for(int i=0; i<intervals.size(); i++) {
            // already insert newInterval
            if(isInsert) {  
                ret.push_back(intervals[i]);
                continue;
            }
            
            // insert newInterval before current interval
            if(newInterval.end < intervals[i].start) {  
                ret.push_back(newInterval);
                ret.push_back(intervals[i]);
                isInsert = true;
                continue;
            }
            
            // combine newInterval with current interval
            if(newInterval.start<=intervals[i].end && intervals[i].start<=newInterval.end) {
                newInterval.start = min(newInterval.start, intervals[i].start);
                newInterval.end = max(newInterval.end, intervals[i].end);
                continue;
            }
            
            // newInterval after current interval
            ret.push_back(intervals[i]);
        }
        
        if(!isInsert) ret.push_back(newInterval);
        return ret;
    }
};

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