3Sum——LeetCode
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)
题意是给定一个数组,找出来三元组之和等0,找出所有的组合并且不重复以字典序输出。
大体思路就是,先给数组排序,然后对数组中的前n-2个元素依次遍历,为啥是前n-2个元素,因为遍历到第n-2个元素时,只剩下第n-1和第n个元素来组成三元组了。
遍历的时候,有left和right两个游标,left就从i+1开始,right就从数组最右length-1开始,如果i、left、right三个元素加起来等0,那么就可以加入结果的List中了,如果<0,那么说明负数的绝对值比正数大,left应该向右移动,反之right向左移动。
注意,因为题目要求重复的组合不输出,那么遍历的时候要注意跳过相同的元素。
Talk is cheap>>
public List<List<Integer>> threeSum(int[] num) { List<List<Integer>> res = new ArrayList<>(); if (num == null || num.length < 3) { return res; } Arrays.sort(num); if (num[0] > 0 || num[num.length - 1] < 0) return res; for (int i = 0; i < num.length - 2; i++) { if (i != 0 && num[i] == num[i - 1]) { continue; } int left = i + 1, right = num.length - 1; while (left < right) { int sum = 0; List<Integer> tmp = new ArrayList<>(); sum = num[i] + num[left] + num[right]; if (sum == 0) { tmp.add(num[i]); tmp.add(num[left]); tmp.add(num[right]); res.add(tmp); while (left < right && num[left] == num[left + 1]) { left++; } while (left < right && num[right] == num[right - 1]) { right--; } left++; right--; } else if (sum < 0) { left++; } else { right--; } } } return res; }