leetcode 18. 4Sum
2017-08-01 本文已影响0人
云胡同学
题目描述
Given an array S of n integers, are there elements a, b, c, and d in S 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.
For example, given array S = [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]
]
思路
目的数组S中找出四个元素使之和为target,与3sum类似,只是外面变为了二重循环。
注意去重
代码
class Solution {
public:
vector<vector <int> > fourSum(vector<int>& nums, int target) {
int i, j;
vector<vector <int>> res;
if (nums.size() < 4)
return res;
sort(nums.begin(), nums.end());
if (nums.size() == 4)
{
if (nums[0] + nums[1] + nums[2] + nums[3] == target)
{
vector<int> ini;
ini.push_back(nums[0]);
ini.push_back(nums[1]);
ini.push_back(nums[2]);
ini.push_back(nums[3]);
res.push_back(ini);
}
else
{
}
return res;
}
for (i = 0; i < nums.size() - 3; i++)
{
for (j = i + 1; j < nums.size() - 2; j++)
{
int k = j + 1;
int l = nums.size() - 1;
while (k < l)
{
if (nums[i] + nums[j] + nums[k] + nums[l] > target)
{
l--;
}
else if (nums[i] + nums[j] + nums[k] + nums[l] < target)
{
k++;
}
else if (nums[i] + nums[j] + nums[k] + nums[l] == target)
{
vector<int> ini;
ini.push_back(nums[i]);
ini.push_back(nums[j]);
ini.push_back(nums[k]);
ini.push_back(nums[l]);
res.push_back(ini);
k++;
l--;
while (nums[k - 1] == nums[k])
k++;
while (nums[l] == nums[l + 1])
l--;
}
}
while (nums[j] == nums[j + 1])
j++;
}
while (nums[i] == nums[i + 1])
i++;
}
return res;
}
};