Problem
Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
The solution set must not contain duplicate triplets.
Example:
Given array nums = [-1, 0, 1, 2, -1, -4],
A solution set is:
[
[-1, 0, 1],
[-1, -1, 2]
]
Solution (Java)
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
final int len = nums.length;
List<List<Integer>> result = new ArrayList<>();
int l;
int r;
for (int i = 0; i < len - 2; i++) {
l = i + 1;
r = len - 1;
while (r > l) {
int sum = nums[i] + nums[l] + nums[r];
if (sum < 0) {
l++;
} else if (sum > 0) {
r--;
} else {
List<Integer> list = new ArrayList<>();
list.add(nums[i]);
list.add(nums[l]);
list.add(nums[r]);
result.add(list);
while (l < r && nums[l + 1] == nums[l]) {
l++;
}
while (r > l && nums[r - 1] == nums[r]) {
r--;
}
l++;
r--;
}
}
while (i < len - 1 && nums[i + 1] == nums[i]) {
i++;
}
}
return result;
}
}
Solution (Javascript)
/**
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function(nums) {
var len = nums.length;
var res = [];
var l = 0;
var r = 0;
nums.sort((a, b) => (a - b));
for (var i = 0; i < len; i++) {
if (i > 0 && nums[i] === nums[i - 1]) continue;
l = i + 1;
r = len - 1;
while (l < r) {
if (nums[i] + nums[l] + nums[r] < 0) {
l++;
} else if (nums[i] + nums[l] + nums[r] > 0) {
r--;
} else {
res.push([nums[i], nums[l], nums[r]]);
while (l < r && nums[l] === nums[l + 1]) l++;
while (l < r && nums[r] === nums[r - 1]) r--;
l++;
r--;
}
}
}
return res;
};
Explain:
先排序，用双指针求解，注意重复的数字。
Complexity:
- Time complexity : O(n^2).
- Space complexity : O(1).