題目：

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].

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思路： 對于intervals從頭到尾遍歷，用new interval與相應的interval[i]對比，如果整個new interval都在interval[i]的左邊，就把new interval存到result里，然后把剩下的所有interval都存起來即可，如果new interval在interval[i]的右邊，則存起來interval[i]，然后繼續遍歷； 如果new interval 和 interval[i]有overlap， 則將兩者進行合并，然后繼續遍歷。

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代碼：?

1 class Solution { 2 public: 3 vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 vector<Interval> result; 7 8 for (int i = 0; i<intervals.size(); i++){ 9 10 if (intervals[i].end < newInterval.start) 11 result.push_back(intervals[i]); 12 else if (intervals[i].start>newInterval.end){ 13 14 result.push_back(newInterval); 15 while (i<intervals.size()){ 16 17 result.push_back(intervals[i]); 18 i++; 19 } 20 } 21 22 else{ 23 24 newInterval.start = min(intervals[i].start, newInterval.start); 25 newInterval.end = max(intervals[i].end, newInterval.end); 26 27 } 28 } 29 30 if(0 == result.size() || result.back().end < newInterval.start) 31 result.push_back(newInterval); 32 33 return result; 34 } 35 };

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轉載于:https://www.cnblogs.com/tanghulu321/archive/2013/05/11/3072348.html

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