[LeetCode] Insert Interval

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]. 

解题思路:

题意为向一个已有的排序区间插入一个新区间,运行必要的合并。使得结构中没有重合的区间。

解法1,先将待插入的区间插入到原来的区间数组中,然后按Merge Interval的办法运行(http://www.kangry.net/blog/?

type=article&article_id=337)。

可是这种时间复杂度为O(nlogn)。没实用到原区间数组已经排好序这个信息。

/**
 * 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) {
        intervals.push_back(newInterval);
        
        std::sort(intervals.begin(), intervals.end(), comp);
        
        vector<Interval> result;
        
        int len = intervals.size();
        for(int i=0; i<len; i++){
            if(result.size() == 0 || !isOver(result[result.size()-1], intervals[i])){
                result.push_back(intervals[i]);
            }else{
                result[result.size()-1].end = max(result[result.size()-1].end, intervals[i].end);
            }
        }
        
        return result;
    }
    
    static bool comp(Interval& interval1, Interval& interval2){
        return interval1.start < interval2.start;
    }
    
    bool isOver(Interval& interval1, Interval& interval2){
        return interval1.start<=interval2.end && interval1.end>=interval2.start;
    }
};
解法2、扫描区间数组,分情况讨论。

(1)若新区间没有插入,且新区间的起始节点小于当前扫描区间的起始节点,那么将新区间插入

(2)若新区间没有插入。且新区间的起始节点大于当前扫描区间,而且当前扫描区间与新区间没有重合,则插入当前扫描区间

(3)若新区间没有插入。且新区间的起始节点大于当前扫描区间。但当前扫描区间与新区间有重合。则将当前扫描区间与新区间合并后插入到结果中

(4)若新区间已经插入,则按Merge Intereval的方法处理

注意到扫描结束后还应该推断新区间是否已经插入。

/**
 * 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) {
        vector<Interval> result;
        
        int len = intervals.size();
        
        bool flag = false;  //newInterval是否增加到了result中了
        for(int i=0; i<len; i++){
            if(!flag){
                if(newInterval.start < intervals[i].start){
                    result.push_back(newInterval);
                    flag = true;
                    i--;
                }else if(!isOver(newInterval, intervals[i])){
                    result.push_back(intervals[i]);
                }else{
                    newInterval.start = min(newInterval.start, intervals[i].start);
                    newInterval.end = max(newInterval.end, intervals[i].end);
                    result.push_back(newInterval);
                    flag = true;
                }
            }else if(!isOver(result[result.size() - 1], intervals[i])){
                result.push_back(intervals[i]);
            }else{
                result[result.size() - 1].end = max(result[result.size() - 1].end, intervals[i].end);
            }
        }
        
        if(!flag){
            if(result.size() == 0 || !isOver(result[result.size() - 1], newInterval)){
                result.push_back(newInterval);
            }else{
                result[result.size() - 1].end = max(result[result.size() - 1].end, newInterval.end);
            }
        }
        
        return result;
    }
    
    bool isOver(Interval& interval1, Interval& interval2){
        return interval1.start<=interval2.end && interval1.end>=interval2.start;
    }
};

posted @ 2017-07-18 19:55  zsychanpin  阅读(127)  评论(0编辑  收藏  举报