LeetCode -- 3Sum

Question：

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)

Analysis：

在sum系列中，这是最后做的一道，用前面的方法却错误多多，很是烦恼。

2sum中，使用了两种方法，暴力求解（2层for循环）和使用Hashmap的方法；

3sum中，使用2个指针分别指向最大和最小值，同时要查重，避免重复的三元组放入list中。

3Sum Closet中，用两个变量记录当前sum和sum与target间的差距，而无需去重检验，总体思路同3Sum。

4sum中，使用两个for循环和2个指针，分别指向最大和最小的值，然后用HashSet纪录选择过的四元组。

Answer:

public class Solution {private List<List<Integer>> res;public List<List<Integer>> threeSum(int[] nums) {res = new ArrayList<List<Integer>> ();Arrays.sort(nums);for(int i=0; i<=nums.length-3; i++) {if(i!=0 && nums[i] == nums[i-1])continue;deal(nums, i, i+1, nums.length-1);}return res;}public void deal(int[] nums, int i, int p, int q) {while(p<q) {if(nums[p] + nums[q] + nums[i] > 0) {q--;}else if(nums[p] + nums[q] + nums[i] < 0) {p++;}else {List<Integer> li = new ArrayList<Integer> ();li.add(nums[i]);li.add(nums[p]);li.add(nums[q]);res.add(li);p++;q--;while(p<q && nums[p]==nums[p-1]) {p++;}while(p<q && nums[q]== nums[q+1]) {q--;}}}}
}

转载于:https://www.cnblogs.com/little-YTMM/p/4789279.html

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