Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
- Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
- The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4}, A solution set is: (-1, 0, 1) (-1, -1, 2)
以一个数为target然后循环求得是否有两个数等于之,while循环里是关键算法。
public class Solution {
public List<List<Integer>> threeSum(int[] num) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(num == null||num.length<3){
return res;
}
Arrays.sort(num);
for(int i=0;i<num.length-2;i++){
int target = -num[i];
int left = i+1;
int right = num.length - 1;
while(left<right){
if(target == num[left]+num[right]){
List<Integer> lists= new ArrayList<Integer>();
lists.add(num[i]);
lists.add(num[left]);
lists.add(num[right]);
if(res.isEmpty()||!res.contains(lists)){
res.add(lists);
}
left++;
right--;
}else if(target > num[left]+num[right]){
left++;
}else{
right--;
}
}
}
return res;
}
}