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15.3sum.cpp
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104 lines (99 loc) · 2.9 KB
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// Tag: Array, Two Pointers, Sorting
// Time: O(N^2)
// Space: O(1)
// Ref: -
// Note: -
// Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
// Notice that the solution set must not contain duplicate triplets.
//
// Example 1:
//
// Input: nums = [-1,0,1,2,-1,-4]
// Output: [[-1,-1,2],[-1,0,1]]
// Explanation:
// nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
// nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
// nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
// The distinct triplets are [-1,0,1] and [-1,-1,2].
// Notice that the order of the output and the order of the triplets does not matter.
//
// Example 2:
//
// Input: nums = [0,1,1]
// Output: []
// Explanation: The only possible triplet does not sum up to 0.
//
// Example 3:
//
// Input: nums = [0,0,0]
// Output: [[0,0,0]]
// Explanation: The only possible triplet sums up to 0.
//
//
// Constraints:
//
// 3 <= nums.length <= 3000
// -105 <= nums[i] <= 105
//
//
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
int n = nums.size();
sort(nums.begin(), nums.end());
vector<vector<int>> res;
for (int k = 2; k < n; k++) {
if (k + 1 < n && nums[k] == nums[k + 1]) {
continue;
}
int target = -nums[k];
int l = 0;
int r = k - 1;
while (l < r) {
if (nums[l] + nums[r] == target) {
res.push_back({nums[l], nums[r], nums[k]});
while (l < r && nums[r] == nums[r - 1]) {
r--;
}
while (l < r && nums[l] == nums[l + 1]) {
l++;
}
l++;
r--;
} else if (nums[l] + nums[r] > target) {
r--;
} else {
l++;
}
}
}
return res;
}
};
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
int n = nums.size();
sort(nums.begin(), nums.end());
vector<vector<int>> res;
for (int i = 0; i < n; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int target = -nums[i];
unordered_map<int, int> history;
for (int j = i + 1; j < n; j++) {
int tmp = target - nums[j];
if (history.count(tmp) > 0) {
res.push_back({nums[i], nums[j], tmp});
while (j + 1 < n && nums[j] == nums[j + 1]) {
j++;
}
} else {
history[nums[j]] = j;
}
}
}
return res;
}
};