leetCode 18.4Sum (4数字和) 解题思路和方法

4Sum 
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)

思路：此题与3Sum是相关的题，此题转换成b+c+d = target - a =k的3Sum。

具体方法和思路见如下代码：

    public List<List<Integer>> fourSum(int[] nums, int target) {        List<List<Integer>> list = new ArrayList<List<Integer>>();        int len = nums.length;        if(len <= 3)//长度不足，直接返回            return list;        Arrays.sort(nums);//排序        //开始循环        for(int i = 0; i < len - 2; i++){        //如果target> 0 且nums[i]>target，则剩余数组相加不可能=target            if((target > 0 && nums[i] > target) || (target < 0 && nums[len-1] < target))                break;                        if(i > 0 && nums[i] == nums[i-1])//消除重复                continue;                        //由a+b+c+d = t 转换成b + c + d = t - a            int a = target - nums[i];            for(int j = i + 1; j < len - 1;j++){                if((a > 0 && nums[j] > a) || (a < 0 && nums[len-1] < a))//如果nums[j]>a，则剩余数组相加不可能=a                    break;                            if(j > i+1 && nums[j] == nums[j-1])                    continue;                                int m = j+1;                int n = len - 1;                                while(m < n){                    int k = nums[j] + nums[m] + nums[n];                    if(k == a){                        List<Integer> al = new ArrayList<Integer>();                        al.add(nums[i]);                        al.add(nums[j]);                        al.add(nums[m]);                        al.add(nums[n]);                        list.add(al);                        m++;                        n--;                        while(m < n && nums[m] == nums[m-1])                            m++;                        while(m < n && nums[n] == nums[n+1])                            n--;                    }                    else{//分情况改变位置标记                        if(k < a)                            m++;                        else                            n--;                    }                }            }        }        return list;    }