[LeetCode]4Sum

最新推荐文章于 2019-01-06 11:55:02 发布

转载最新推荐文章于 2019-01-06 11:55:02 发布 · 500 阅读

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本文深入探讨了如何在一个整数数组中寻找四个元素的唯一组合，使得它们的和等于指定的目标值。文章详细介绍了解题思路、算法实现以及去重策略，确保结果的正确性和唯一性。

4Sum

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Total Accepted: 50999 Total Submissions: 230927 Difficulty: Medium

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.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.

    A solution set is:
    (-1,  0, 0, 1)
    (-2, -1, 1, 2)
    (-2,  0, 0, 2)

这个题目跟3Sum的题目是一样的，只不过多了一个循环，去重也多了一步，对于第二个数也不能跟它之前的数重复。在找到一个解之后也要继续去重。

public class Solution {
    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        ArrayList<List<Integer>>result = new ArrayList<List<Integer>>();
        if(nums.length<4)return result;        
        for(int i = 0;i<=nums.length-4;i++){
        	if(i!=0&&nums[i]==nums[i-1]){
        		continue;
        	}
        	for(int j =i+1;j<=nums.length-3;j++){
        		if(j!=i+1&&nums[j]==nums[j-1]){
        			continue;
        		}
        		int p = j+1;int q = nums.length-1;
        		while(p<q){
        			int sumTemp = nums[i]+nums[j]+nums[p]+nums[q];
        			if(sumTemp==target){
            			ArrayList<Integer>sumArrayList = new ArrayList<Integer>();
            	        sumArrayList.add(nums[i]);sumArrayList.add(nums[j]);
            	        sumArrayList.add(nums[p]);sumArrayList.add(nums[q]);
            	        result.add(sumArrayList);
            	        p++;
            	        while(p<q&&nums[p]==nums[p-1]){
            	        	p++;
            	        }
            	        q--;
            	        while(p<q&&nums[q]==nums[q+1]){
            	        	q--;
            	        }
            		}
            		else{
            			if(sumTemp>target){
            				q--;
            			}
            			else{
            				p++;
            			}
            		}
        		}
        		
        	}
        }
        return result;
    }
}