4Sum
4Sum
问题：
Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
- Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
- The solution set must not contain duplicate quadruplets.
思路：
双指针问题的扩展
我的代码：
public class Solution { public List<List<Integer>> fourSum(int[] num, int target) { List<List<Integer>> list = new ArrayList<List<Integer>>(); if(num == null || num.length < 4) return list; int len = num.length; Arrays.sort(num); for(int i = 0; i <= len - 4; i++) { if(i == 0 || !(i != 0 && num[i] == num[i-1])) { for(int j = i + 1; j <= len - 3; j++) { if(j == i + 1 || !(j != i+ 1 && num[j] == num[j-1])) { int left = j + 1; int right = len - 1; while(left < right) { int val = num[i] + num[j] + num[left] + num[right]; if(val == target) { List<Integer> tmpList = new ArrayList<Integer>(); tmpList.add(num[i]); tmpList.add(num[j]); tmpList.add(num[left]); tmpList.add(num[right]); list.add(tmpList); left++; right--; while(left < right && num[left] == num[left-1]) { left++; } while(left < right && num[right] == num[right+1]) { right--; } } else if(val > target) { right --; } else left ++; } } } } } return list; } }
学习之处：
- 避免重复的关键之处在于：
while(left < right && num[left] == num[left-1]) { left++; }