Partition to K Equal Sum Subsets——LeetCode进阶路

原题链接https://leetcode.com/problems/partition-to-k-equal-sum-subsets/

题目描述

Given an array of integers nums and a positive integer k, find whether it’s possible to divide this array into k non-empty subsets whose sums are all equal.

Example 1:

Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4
Output: True
Explanation: It’s possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.

Note:

1 <= k <= len(nums) <= 16.
0 < nums[i] < 10000.

思路分析

给定一个整数数组，一个整数，返回是否数组能够被分为k个等值的子集。
直观的想，用递归，一个一个子集的凑，一旦有凑不上的则返回false
还有就是注意些特殊情况：

• 分组数大于数组的长度，返回false
• 若数组元素之和不能被数组数整除，则肯定不能满足分组，返回false
• 若分组数为1的话，直接返回数组本身就好，显然返回true
• 当有超过平均值的数组元素，返回false
  notes : 一定记得回溯……
  笔者小陌第N次忘记回溯(╯︵╰)

AC解

class Solution {
    public boolean canPartitionKSubsets(int[] nums, int k) {
        if(k == 1)
        {
            return true;
        }

        if(nums == null || nums.length == 0 || k > nums.length)
        {
            return false;
        }

        int sum = 0;
        int max = 0;
        for(int i:nums)
        {
            sum = sum + i ;
            max = Math.max(max,i);
        }
        if(sum % k != 0 || sum / k < max)
        {
            return false;
        }

        boolean[] flag = new boolean[nums.length];
        return dfs(nums,flag,k,sum /k,0,0);
    }

    public boolean dfs(int[] nums , boolean[] flag , int k , int subset , int curSubset , int position)
    {
       if(k == 0)
       {
           return true;
       }

        if(subset == curSubset)
        {
            return dfs(nums,flag,k-1,subset,0,0);
        }

        for(int i = position ; i<nums.length ; i++)
        {
            if(flag[i])
            {
                continue;
            }
            flag[i] = true;
            if(dfs(nums,flag,k,subset,curSubset+nums[i],i+1))
            {
                return true;
            }
            flag[i] = false;//记得回溯！！！！
        }

        return false;
    }
}