结构数组新发现之直接初始化《leetcode-合并区间》

leetcode有时候会要求一些奇怪（陌生）的数据形式，刷题因为数据形式卡住了真的很不好...

合并区间里定义了一个Interval的结构数组

struct Interval {
    int start;
    int end;
    Interval() : start(), end() {}
    Interval(int s, int e) : start(s), end(e) {}
};

又根据结构数组创建了vector<Interval> intervals。我就在想，怎么初始化这个intervals。

vector<Interval> intervals;
Interval int1(,); intervals.push_back(int1);
Interval int2(,); intervals.push_back(int2);

其实就是利用结构数组里的定义来，这样初始化真的很方便呐，也不用去特意申请空间。

解题思路：

先根据start排序，再一一合并

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <vector>
using namespace std;

typedef struct Interval Int;
struct Interval {
    int start;
    int end;
    Interval() : start(), end() {}
    Interval(int s, int e) : start(s), end(e) {}
};
class Solution {
public:
    vector<Interval> merge(vector<Interval>& intervals) {
        int n = intervals.size();
        ) return intervals;
        int j;
        Interval temp;
        Interval temp0;
        ;
        ;i<n;i++){
            temp = intervals[i];
             && temp.start<intervals[j-].start;j--){
                intervals[j].start = intervals[j-].start;
                intervals[j].end = intervals[j-].end;
            }
            intervals[j].start = temp.start;
            intervals[j].end = temp.end;
        }
        vector<Interval> outcome;
        outcome.push_back(intervals[]);
        ;i<n;i++){
            if(intervals[i].start <= outcome.back().end){
                outcome.back().end = outcome.back().end > intervals[i].end ? outcome.back().end:intervals[i].end;
            }
            else{
                outcome.push_back(intervals[i]);
            }
        }
        return outcome;
    }
};

int main(){
    vector<Interval> intervals;
    Interval int1(,); intervals.push_back(int1);
    Interval int2(,); intervals.push_back(int2);
//    Interval int3(8,10); intervals.push_back(int3);
//    Interval int4(15,18); intervals.push_back(int4);

    Solution S;
    vector<Interval> outCome = S.merge(intervals);
    ;i<outCome.size();i++){
        int1 = outCome[i];
        printf("%d %d\n", int1.start, int1.end);
    }
    ;

}