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64 lines (53 loc) · 2.02 KB
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/* 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. */
class ThreeSum {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> result = new LinkedList<>();
for(int i=0;i<nums.length-2;i++){
if(i==0 || (i>0 && nums[i]!=nums[i-1])){
int left = i+1;
int right = nums.length-1;
int sum = 0-nums[i];
while(left < right){
if(nums[left]+nums[right] == sum){
result.add(Arrays.asList(nums[i],nums[left],nums[right]));
while(left < right && nums[left]==nums[left+1]){
left++;
}
while(left < right && nums[right] == nums[right-1]){
right--;
}
left++;
right--;
}
else if(nums[left] + nums[right] < sum){
left++;
}
else {
right--;
}
}
}
}
return result;
}
}