[算法分析与设计] leetcode 每周一题: Find Right Interval

题目链接：https://leetcode.com/problems/find-right-interval/description/

题目大意：给定一组区间（i, j），找出每一个区间“右边”最近的区间的索引

思路：

将题目给的interval 组转化为带有索引的indexIntreval组， 然后排序，接着对每个区间，遍历其右边的区间比较

代码：

/** * 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:    struct indexInterval {        int start;        int end;        int index;        indexInterval():indexInterval(-1,-1,-1){}        indexInterval(int s, int e, int i): start(s), end(e), index(i) {}    };    bool static compare(indexInterval a, indexInterval b) {        return a.start <= b.start;    }    vector<int> findRightInterval(vector<Interval>& intervals) {        vector<indexInterval> localSet;        int len = intervals.size();        for(int i = 0; i < len; i++) {            int start = intervals[i].start;            int end = intervals[i].end;            localSet.push_back(indexInterval(start, end, i));                    }        sort(localSet.begin(), localSet.end(), compare);        vector<int> result(len,-1);        for(int i = 0; i < len; i++) {            int target = -1;            for(int j = i + 1; j < len; j++) {                                if(localSet[j].start >= localSet[i].end) {                    target = localSet[j].index;                    break;                }            }            result[localSet[i].index] = target;            // target = -1;        }        return result;    }};