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Description
题目描述
给定一个包含 n 个整数的数组 nums 和一个目标值 target,判断 nums 中是否存在四个元素 a,b,c 和 d ,使得 a + b + c + d 的值与 target 相等?找出所有满足条件且不重复的四元组。
注意:
答案中不可以包含重复的四元组。
示例:
给定数组 nums = [1, 0, -1, 0, -2, 2],和 target = 0。
满足要求的四元组集合为:
[
[-1, 0, 0, 1],
[-2, -1, 1, 2],
[-2, 0, 0, 2]
]
链接:https://leetcode-cn.com/problems/4sum
方案
固定前两个数,后两个数用双指针法。此题与三个数的和类似#7。
class Solution {
public:
void fastsort(vector<int>& nums, int a, int b){
if (a >= b || nums.size() <= 1)return;
int i, j, tmp = nums[a];
for (i = a, j = b; i < j;){
while (i<j && nums[j] >= tmp)j--;
nums[i] = nums[j];
while (i < j&&nums[i] <= tmp)i++;
nums[j] = nums[i];
}
nums[j] = tmp;
fastsort(nums, a, j - 1);
fastsort(nums, j + 1, b);
}
vector<vector<int>> fourSum(vector<int>& nums, int target) {
fastsort(nums, 0, nums.size() - 1);
int a, b, c, d;
int len = nums.size();
vector<vector<int>> res = {};
for (a = 0; a + 3<len;){
for (b = a + 1; b + 2<len;){
c = b + 1;
d = len - 1;
while (c<d){
if ((nums[a] + nums[b] + nums[c] + nums[d])<target){
c++;
}
else if ((nums[a] + nums[b] + nums[c] + nums[d])>target){
d--;
}
else{
res.push_back({ nums[a], nums[b], nums[c], nums[d] });
int ctmp = nums[c], dtmp = nums[d];
do{
c++;
} while (c<d&&nums[c] == ctmp);
do{
d--;
} while (c<d&&nums[d] == dtmp);
}
}
do{
b++;
} while (b + 2<len&&nums[b]==nums[b-1]);
}
do{
a++;
} while (a + 3<len&&nums[a]==nums[a-1]);
}
return res;
}
};
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