56[LeetCode] .Merge Intervals

时间:2023-12-25 09:35:43

Given an array nums of n integers and an integer target, are there elements abc, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]



注意sort() 中的cmp()比较函数的定义要放在类外面:

/**
* Definition for an interval.
* struct Interval {
* int start;
* int end;
* Interval() : start(0), end(0) {}
* Interval(int s, int e) : start(s), end(e) {}
* };
*/
bool cmp(Interval a,Interval b){return a.start<b.start;}
class Solution {
public: vector<Interval> merge(vector<Interval>& ins) {
if (ins.empty()) return vector<Interval>{};
vector<Interval> res;
sort(ins.begin(), ins.end(), cmp);
res.push_back(ins[]);
for (int i = ; i < ins.size(); i++) {
if (res.back().end < ins[i].start) res.push_back(ins[i]);
else
res.back().end = max(res.back().end, ins[i].end);
}
return res;
}
};

如在在sort中定义排序方法应该这么写:

vector<Interval> merge(vector<Interval>& ins) {
if (ins.empty()) return vector<Interval>{};
vector<Interval> res;
sort(ins.begin(), ins.end(), [](Interval a, Interval b){return a.start < b.start;});
res.push_back(ins[]);
for (int i = ; i < ins.size(); i++) {
if (res.back().end < ins[i].start) res.push_back(ins[i]);
else
res.back().end = max(res.back().end, ins[i].end);
}
return res;
}