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4Sum.cc
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66 lines (55 loc) · 2.39 KB
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// https://oj.leetcode.com/problems/4sum/
// 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:
// Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
// 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)
namespace FourSum {
class Solution {
public:
vector<vector<int> > fourSum(vector<int> &num, int target) {
vector<vector<int> > res;
if (num.size() < 4) {
return res;
}
sort(num.begin(), num.end());
for (int i = 0; i < num.size() - 3; i++) {
if (i != 0 && num[i] == num[i - 1]) continue; // skip duplication
for (int j = i + 1; j < num.size() - 2; j++) {
if (j != i + 1 && num[j] == num[j - 1]) continue; // skip duplication
int m = j + 1;
int n = num.size() - 1;
while (m < n) {
if (num[i] + num[j] + num[m] + num[n] == target) {
vector<int> curRes;
curRes.push_back(num[i]);
curRes.push_back(num[j]);
curRes.push_back(num[m]);
curRes.push_back(num[n]);
res.push_back(curRes);
do {
m++;
} while(m < n && num[m] == num[m - 1]); // skip duplication
do {
n--;
} while(m < n && num[n] == num[n + 1]); // skip duplication
} else if (num[i] + num[j] + num[m] + num[n] > target) {
do {
n--;
} while(m < n && num[n] == num[n + 1]); // skip duplication
} else {
do {
m++;
} while(m < n && num[m] == num[m - 1]); // skip duplication
}
}
}
}
return res;
}
};
}