57. Insert Interval

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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].

分析

把插入newInterval看成如下形式，想象成把一个长方形的容器倒盖在现有的排好顺序的容器上
  |~~~~~~~~~~~~~~~~~~~~~~|       newInterval
|____| |__| |________|  |______| intervals

先确定newInterval的 start 在哪个位置，假设当前指向的intervals[i] 叫做 curi
如果newInterval.end > curi.end
那么将该curi 压入结果 result，并继续对下一个interval进行判断(可能会出现newInterval超出边界的情况，则直接压入结果result的最末端)

否则：
1 在curi的内部，即满足 newInterval.start >= curi.start 
     |~~~~~~~~~~~~~~......... newInterval 
...|____| |__| |_______...... intervals
2 在curi的前面的空隙，即满足 newInterval.start < curi.start
      |~~~~~~~~~~~~~~~~~~......... newInterval 
..._|   |____| |__| |_______...... intervals
确定新interval，即newi的start，只需要取 curi.start 和 newInterval.start的最小值就行

先给newi的end赋值 newInterval.end，然后不断跳过那些在 newInterval.end 范围内的 intervals[i], 直到curi.end >= newInterval.end，如下图所示两种情况：
.....~~~~~~~~~~~~~~~~~~~~|       newInterval
..|____| |__| |________|  |______| intervals
                            curi

.....~~~~~~~~~~~~~~~~~~~~~~~~~|    newInterval
..|____| |__| |________|  |______| intervals
                            curi
分别对这两种情况确定 newi.end

最后将剩余的intervals[i] 压入结果result

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/**
 * 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) {
        vector<Interval> result;
        int i;
        for(i = 0; i < intervals.size(); ++i){
            if(intervals[i].end < newInterval.start){
                result.push_back(intervals[i]);
            }
            else{
                break;
            }
        }
        // new interval is out of the max range
        if(i == intervals.size()){
            result.push_back(newInterval);
            return result;
        }
         
        Interval newi(min(intervals[i].start, newInterval.start), newInterval.end);
        for(; i < intervals.size(); ++i){
            Interval curi = intervals[i];
            if(curi.end < newi.end){
                continue;
            }
            else{
                // newi end is int the range of curi
                if(newi.end >= curi.start){
                    newi.end = curi.end;
                    ++i; break;
                }
                //newi end is before the rnage of curi
                else{
                    break;
                }
            }
        }
        result.push_back(newi);
         
        for(;i < intervals.size(); ++i){
            result.push_back(intervals[i]);
        }
        return result;
    }
};




posted @ 2017-02-14 21:46  copperface  阅读(240)  评论(0)  编辑  收藏  举报