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56[LeetCode] .Merge Intervals
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Given an array
nums
of
n
integers and an integer
target
, are there elements
a
,
b
,
c
, 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;
}
