leetcode 18. 4Sum

leetcode 18. 4Sum 题目，但是其解题思路可以延伸到 ksum

package com.protobuf.ACM;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

/**
 * Created by liuchaoqun01 on 2019/2/19.
 */
public class Ksum {
    /**
     * 对于此类问题,一般的解决思路是先排序. 然后将前k-2个数字进行循环,在剩下的范围内设置两个指针p,q.
     * 遍历p,q的值, 直到达到target
     */
    public static List<List<Integer>> fourSum(int[] nums, int target) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);
        for (int i = 0; i < nums.length - 3; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            for (int j = i + 1; j < nums.length - 2; ++j) {
                // 这里j一定是大于i + 1
                if (j > i + 1 && nums[j] == nums[j - 1]) {continue;}
                int left = j + 1;
                int right = nums.length - 1;
                while (left < right) {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    // 命中以后再来判断重复
                    if (sum == target) {
                        List<Integer> tmp = Arrays.asList(nums[i], nums[j], nums[left], nums[right]);
                        res.add(tmp);
                        // 避免重复值
                        while (left < right && nums[left] == nums[left + 1]) {
                            ++left;
                        }
                        while (left < right && nums[right] == nums[right - 1]) {
                            --right;
                        }
                        ++left;
                        --right;
                    } else if (sum < target) {
                        ++ left;
                    } else {
                        -- right;
                    }
                }
            }
        }
        return res;
    }

    public static void main(String[] args) {
        int[] nums = {0,0,0,0};
        int target = 0;
        List<List<Integer>> res = Ksum.fourSum(nums, target);
        System.out.println("ok");
    }
}

参考：
1 ksum