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leetcode_0018_4Sum.cpp
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58 lines (53 loc) · 1.62 KB
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class Solution {
public:
/**
* AC:
* Runtime: 48 ms, faster than 30.52% of C++ online submissions for 4Sum.
* Memory Usage: 10.6 MB, less than 11.68% of C++ online submissions for 4Sum.
*
* 思路：先排序，然后以 O(n^2) 的复杂度做遍历，每一轮用两个游标缩小范围。
* 总体复杂度：O(n^2)
*/
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int> >ret;
int size = nums.size();
if(size < 4)
return vector<vector<int> >();
// sort
sort(nums.begin(), nums.end());
for(int i = 0; i < size - 1; i++) {
for(int j = i + 1; j < size; j++) {
int l = j + 1;
int r = size - 1;
while(l < r) {
int tempSum = nums[i] + nums[j] + nums[l] + nums[r];
if(tempSum == target) {
vector<int>tempArr;
tempArr.push_back(nums[i]);
tempArr.push_back(nums[j]);
tempArr.push_back(nums[l]);
tempArr.push_back(nums[r]);
int flag = 0;
for(auto it:ret) {
if(it == tempArr) {
flag = 1;
break;
}
}
if(flag) {
l++;
continue;
}
ret.push_back(tempArr);
// break; // 这里不应该直接退出，因为第三四个有可能有多种满足的组合
l++;
} else if(tempSum > target)
r--;
else if(tempSum < target)
l++;
}
}
}
return ret;
}
};