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)

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Have you met this question in a real interview?

思路 ： 和 3sum 一樣 ， 多加一重循環而已， 

易錯點： 注意去除重複。還有 找到一個解法以後的left＋＋， right －－

public class Solution {
    public List<List<Integer>> fourSum(int[] num, int target) {
        List<List<Integer>> ret = new ArrayList<List<Integer>>();
        int len = num.length;
        Arrays.sort(num);
        
        for(int i=0; i<len; i++) {
            if(i >= 1 && i < len && num[i] == num[i-1]) {//------
                continue;
            }
            for(int j=i+1; j<len; j++) {
                if(j >= i+2 && j < len && num[j] == num[j-1]) {//------
                    continue;
                }
                int left = j+1, right = len-1;
                while(left < right) {
                    int sum = num[i] + num[j] + num[left] + num[right];
                    if(sum == target) {
                        List<Integer> al = new ArrayList<Integer>();
                        al.add(num[i]);
                        al.add(num[j]);
                        al.add(num[left]);
                        al.add(num[right]);
                        ret.add(al);
                        left++;//--
                        right--;//--
                        while(left < right && num[left] == num[left-1]){//----
                            left++;
                        }
                        while(left < right && num[right] == num[right+1]){//----
                            right--;
                        }
                    } else if(sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }
        return ret;
    }
}