leetcode 18 4Sum

leetcode 18 4Sum

原题链接：https://leetcode.com/problems/4sum/

Description

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]
]

Solution

二分查找,先预先枚举出a+b所得的n2个数字并排好序。

然后判断是否有a+b=target−c−d(二分查找即可)

PS: 注意所求的四个数下标各不相同,枚举的时候要加以判断

PPS: 所得结果可能需要去重。

class Solution {
using vec = vector<int>;
using mat = vector<vec>;
using pii = pair<int, pair<int, int>>;
public:
mat fourSum(vec& nums, int target) {
mat res;
int n = 0;
if(!(n = nums.size()) || n < 4) return res;
vector<pii> vpi;
for(int i = 0; i < n; i++) {
for(int j = i + 1; j < n; j++) {
vpi.push_back({ nums[i] + nums[j], { i, j } });
}
}
sort(vpi.begin(), vpi.end(), [](pii &a, pii &b)->bool {
return a.first < b.first;
});
int size = vpi.size();
for(int i = 0; i < n; i++) {
for(int j = i + 1; j < n; j++) {
int lb = 0, ub = size - 1;
int k = target - nums[i] - nums[j];
while(lb <= ub) {
int m = (lb + ub) >> 1;
if(vpi[m].first >= k) ub = m - 1;
else lb = m + 1;
}
while(lb < size && vpi[lb].first == k) {
auto &r = vpi[lb++].second;
if(r.first == i || r.first == j || r.second == i || r.second == j) continue;
int a = nums[i], b = nums[j], c = nums[r.first], d = nums[r.second];
if(a + b + c + d != target) continue;
if (a > b) swap(a, b);
if (c > d) swap(c, d);
if (a > c) swap(a, c);
if (b > d) swap(b, d);
if (b > c) swap(b, c);
res.push_back({ a, b, c, d });
}
}
}
sort(res.begin(), res.end());
res.erase(unique(res.begin(), res.end()), res.end());
return res;
}
};