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program59.java
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102 lines (68 loc) · 2.56 KB
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/*
15. 3Sum
Medium
25.5K
2.3K
Companies
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.
*/
package LeetCode;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
public class program59 {
static List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>>ls = 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 sum = 0-nums[i];
int right = nums.length-1;
while(left <right){
if(nums[left]+nums[right]==sum){
ls.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 ls;
}
public static void main(String[] args) {
int[] nums = {-1,0,1,2,-1,-4};
List<List<Integer>>ls = threeSum(nums);
System.out.println(ls);
}
}