698. Partition to K Equal Sum Subsets-优快云博客

本文介绍了一个LeetCode上的经典问题——如何将一组数分区为k个子集，每个子集的和相等。通过使用深度优先搜索(DFS)的方法来解决这个问题，并详细解释了状态转移的过程。此外，还提出了一种优化方案：先对数组进行降序排序，优先选择较大的数值放入当前组中。

https://leetcode.com/problems/partition-to-k-equal-sum-subsets/description/

Use DFS. We need to choose k group of numbers (each group sum to s). The state is 1) how many groups left, 2) current sum in current group, 3) visisted for each number.

If given state1 (k, cur, visited), we choose a number (make sure this number is not visited, this number + cur <= s), then the state is updated to (k, cur+current number, visited). Then we need to pick another number in new state. That's the sub-problem.

Optimization: sort nums in desc order, and always pick the first (biggest) number for the first group.

class Solution {
public:
    bool canPartitionKSubsets(vector<int>& nums, int k) {
        int s = 0;
        for (auto i : nums)
            s += i;
        if (s % k != 0) return false;
        s = s / k;
        
        // sort in desc order
        sort(nums.begin(), nums.end());
        reverse(nums.begin(), nums.end());
        
        // first set contains the biggest number
        vector<bool> visited(nums.size(), false);
        visited[0] = true;
        
        return dfs(nums, visited, s, nums[0], k);
    }
    bool dfs(vector<int>& nums, vector<bool>& visited, int s, int cur, int k) {
        if (cur == s) {
            k--;
            cur = 0;
            if (k == 0) return true;
        }
        for (int i = 1; i < nums.size(); i++) {
            if (!visited[i] && nums[i] + cur <= s) {
                visited[i] = true;
                bool res = dfs(nums, visited, s, cur + nums[i], k);
                visited[i] = false;
                if (res)    return res;
            }
        }
        return false;
    }
};