15. 3Sum
- tags
- two pointers
- source
- leetcode
Solution - two pointers
- sort to avoid duplicate
- fix one element, becomes the two sum problem, use two pointers
Edge Cases
- might be multiple results in one two sum
- multiple results may be same, input:
[-2,0,0,2,2]
solution
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> res = new LinkedList<>();
for (int i = 0; i < nums.length - 2; i++) {
if (nums[i] > 0) return res;
if (i != 0 && nums[i] == nums[i - 1]) continue;
int target = 0 - nums[i];
int l = i + 1, r = nums.length - 1;
while (l < r) {
int sum = nums[l] + nums[r];
if (sum == target) {
res.add(Arrays.asList(nums[i], nums[l], nums[r]));
while (l < r && nums[l] == nums[++l]);
while (l < r && nums[r] == nums[--r]);
} else if (sum < target) {
l++;
} else {
r--;
}
}
}
return res;
}
}
Complexity