标签：答案 solution unique continue cat 遍历 return etc for
【题目】
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] ]
【思路】
sort后，头尾两个指针。
sum=0,left++,right--,继续遍历，向中间逼近。
sum>0,right--，需要更小的数使之满足。
sum<0,left++，需要更大的数使之满足。
if(sum==0){ data.add(Arrays.asList(nums[i],nums[left], nums[right])); left++; right--; while(left<right&&nums[left]==nums[left-1]) left++; while(left<right&&nums[right]==nums[right+1]) right--; } else if(left<right&&sum>0) right--; else left++; }
答案去重
if(i>0&&nums[i]==nums[i-1]) continue; while(left<right&&nums[left]==nums[left-1]) left++; while(left<right&&nums[right]==nums[right+1]) right--;
【代码】
class Solution { public List<List<Integer>> threeSum(int[] nums) { List<List<Integer>> data=new ArrayList<>(); Arrays.sort(nums); for(int i=0;i<nums.length-2;i++){ int left=i+1; int right=nums.length-1; if(i>0&&nums[i]==nums[i-1]){ continue;} while(left<right){ int sum=nums[left]+nums[right]+nums[i]; if(sum==0){ data.add(Arrays.asList(nums[i],nums[left], nums[right])); left++; right--; while(left<right&&nums[left]==nums[left-1]) left++; while(left<right&&nums[right]==nums[right+1]) right--; } else if(left<right&&sum>0) right--; else left++; } } return data; } }
标签：答案 solution unique continue cat 遍历 return etc for
原文地址：https://www.cnblogs.com/inku/p/9955638.html
