4Sum 找4个数之和为target@LeetCode

模板化了，要计算 k-sum 复杂度为O(N^(k-1))

package Level3;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

/**
 * 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)
 */
public class S18 {

	public static void main(String[] args) {
		int[] num = {1, 0, -1, 0, -2, 2};
		System.out.println(fourSum(num, 0));
	}
	
	// O(n^3) 
	public static ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
		Set<ArrayList<Integer>> ret = new HashSet<ArrayList<Integer>>();
		Arrays.sort(num);

		for (int i = 0; i < num.length; i++) {
			for (int j = i+1; j < num.length; j++) {
				int m = j+1, n = num.length-1;
				while(m < n){
					if(i!=j && j!=m && m!=n){
						if(num[m] + num[n] == target-num[i]-num[j]){
							ArrayList<Integer> list = new ArrayList<Integer>();
							list.addAll(Arrays.asList(num[i], num[j], num[m], num[n]));
							ret.add(list);
							m++;
							n--;
						}else if(num[m] + num[n] < target-num[i]-num[j]){
							m++;
						}else{
							n--;
						}
					}
				}
			}
		}
		
		return new ArrayList<ArrayList<Integer>>(ret);
	}
	
	// O(n^3 * log(n)) 超时！
	public static ArrayList<ArrayList<Integer>> fourSum2(int[] num, int target) {
		Set<ArrayList<Integer>> ret = new HashSet<ArrayList<Integer>>();
		Arrays.sort(num);

		for (int i = 0; i < num.length; i++) {
			for (int j = num.length-1; j > i+1; j--) {
				for (int k = i+1; k < j; k++) {
					if(i!=j && j!=k){
						int remain = target - (num[i] + num[j] + num[k]);
						int index = Arrays.binarySearch(num, k + 1, j, remain);
						if (index >= 0) {
							ArrayList<Integer> list = new ArrayList<Integer>();
							list.add(num[i]);
							list.add(num[k]);
							list.add(num[index]);
							list.add(num[j]);
							ret.add(list);
						}
					}
				}
			}
		}

		return new ArrayList<ArrayList<Integer>>(ret);
    }

}

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);
                        do {
                            left++;
                        } while(left < right && num[left] == num[left-1]);
                        do {
                            right--;
                        } while(left < right && num[right] == num[right+1]);
                    } else if(sum < target) {
                        left++;
                    } else {
                        right--;
                    }
                }
            }
        }
        return ret;
    }
}