LeetCode.18(454) 4Sum && II

题目18：

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: 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]]

分析：

public class Solution {    public List<List<Integer>> fourSum(int[] nums, int target) {        //用两个循环控制（内循环、外循环）        //one at the end of the outer loop (i-loop)        // the other at the end of the inner loop (j-loop)                //the function return immediately when nums[i]*4 > target        //the inner loop break immediately when nums[j]*4 < target.                List<List<Integer>> list=new ArrayList<List<Integer>>();        Arrays.sort(nums);        int len=nums.length;        for(int i=0;i<len-3;i++){            if(nums[i]*4>target) return list;            for(int j=len-1;j>i+2;j--){                if(nums[j]*4<target)break;                int remain=target-nums[i]-nums[j];                int low=i+1,high=j-1;                while(low<high){                    int sum=nums[low]+nums[high];                    if(sum>remain)high--;//如果比剩下的大，坐移                    else if(sum<remain)low++;//如果比剩下的小,右移                    else{                        //相同                        list.add(Arrays.asList(nums[i],nums[low],nums[high],nums[j]));//将该组数据存入list                        //在内循环寻找相同的解                        while(++low<=high&&nums[low]==nums[low-1])continue;                        while(--high>=low&&nums[high]==nums[high+1])continue;                    }                }                while(j>=1&&nums[j]==nums[j-1])--j;//跳出内循环            }            while(i<len-1&&nums[i]==nums[i+1])++i;//跳出外循环        }        return list;    }}

题目454:

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input:A = [ 1, 2]B = [-2,-1]C = [-1, 2]D = [ 0, 2]Output:2Explanation:The two tuples are:1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 02. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

分析：

class Solution {    public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {        //给定四个长度相同的数组，返回由他们四个组成之和为0的组合数。该题不同于4Sum的在于，每个元素的解限定了范围        //思路:使用HashMap实现，先将A和B的组合方法，之后再查找C和D的结果是否存在它们俩的绝对值和，可能存在多对        //只需要将和累加即可                //注意：暴力解法肯定TLM,该题不考虑统一数组中重复数据的影响        HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();        int count=0;        for(int i=0;i<A.length;i++){            for(int j=0;j<B.length;j++){                int sum=A[i]+B[j];                hm.put(sum,hm.getOrDefault(sum,0)+1);            }        }                //查找匹配        for(int i=0;i<C.length;i++){            for(int j=0;j<D.length;j++){                int sum=C[i]+D[j];                count+=hm.getOrDefault(-1*sum,0);            }        }                return count;    }}