[LeetCode 410] Split Array Largest Sum

最新推荐文章于 2025-02-03 15:37:10 发布

ExcitedZhang

最新推荐文章于 2025-02-03 15:37:10 发布

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分类专栏： LeetCode 文章标签： LeetCoe 动态规划数组

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博客围绕将非负整数数组分割成m个非空连续子数组，使这些子数组中最大和最小的问题展开。指出深搜会超时，应使用动态规划。定义dp[i][j]表示nums[0:j]数组分割i次后的最小和，并给出状态转移方程，还给出了运行效率。

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Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.

Note:
If n is the length of array, assume the following constraints are satisfied:

1 ≤ n ≤ 1000
1 ≤ m ≤ min(50, n)

Examples:

Input:nums = [7,2,5,10,8]  m = 2
Output:18
Explanation:
There are four ways to split nums into two subarrays.The best way is to split it into [7,2,5] and [10,8],
where the largest sum among the two subarrays is only 18.

分析

这道题如果用深搜的话一定会超时的。所以只能使用动态规划。动态规划怎么确定数组的含义就比较重要了。

我们使用dp[i][j]表示nums[0:j]的数组被分割i 次后所得到的最小的sum。那么可以推导出：

dp[i][j] = min(max(dp[i-1][k], sum[k:m])), i <= k < j。

即我们计算i次分割的结果时，我们先去找i-1次分割的结果，然后计算i-1次分割的结果与sum[k:j]的最大值，最后遍历 i <= k < j时的最小值。

Code

class Solution {
public:
    int splitArray(vector<int>& nums, int m) {
        int len = nums.size();
        if (len == 0)
            return 0;
        
        vector<vector<long long>> dp(m, vector<long long>(len, INT_MAX));        
        dp[0][0] = nums[0];
        for (int i = 1; i < len; i ++)
        {
            dp[0][i] = dp[0][i-1] + nums[i];
        }
        
        for (int i = 1; i < m; i ++)
        {
            for (int j = i; j < len; j ++)
            {
                for (int k = i-1; k < j; k ++)
                {
                    dp[i][j] = min(dp[i][j], max(dp[i-1][k], dp[0][j] - dp[0][k]));
                }
            }
        }
        return dp[m-1][len-1];
                                   
    }
};

运行效率

Runtime: 132 ms, faster than 27.65% of C++ online submissions for Split Array Largest Sum.

Memory Usage: 9.2 MB, less than 28.31% of C++ online submissions for Split Array Largest Sum.