算法练习(11) —— Split Array Largest Sum
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算法练习(11) —— Split Array Largest Sum
习题
本题取自 leetcode 中的 Dynamic Programming
栏目中的第410题:
Split Array Largest Sum
题目如下:
Description:
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.
Example:
Input:
nums = [7,2,5,10,8]
m = 2Output:
18Explanation:
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.Note
If n is the length of array, assume the following constraints are satisfied:
- 1 ≤ n ≤ 1000
- 1 ≤ m ≤ min(50, n)
思路与代码
- 先理解一下题意,这次的有点绕: 给你两个输入,一个是整数数组nums,另一个是m。将nums分成m个小数组后,求出每个小数组内所有数的和,再在所有和中找到最大值res,作为输出。题目要让我们找到res的最小值。
- 注意几个点:第一是数组内数字的顺序不能改变,第二是子数组不能为空,至少要有一个数据。
- 思路的话,很容易就想到动态规划去了,毕竟这种类型的题目很适合动态规划。具体的状态转移方程为:
// dp[a][b] : a为当前数组的长度, b为剩余切割的次数 // Xi为切割点 dp[len][m] = min {max(dp[Xi][m-1], dp[len-Xi][m-1])} dp[1][m] = 1 dp[len][0] = len
- 自己当时图方便用递归写的,写的挺粗,感觉运行速度有点慢。可以作以下改进:一是记录dp表,如果递归的时候能找到非0值就可以直接返回;二是将递归化为非递归
具体代码如下:
#include <iostream>#include <vector>using namespace std;class Solution {public: int split(vector<int>& nums, int m) { int len = nums.size(); if (len == 1) return nums[0]; /*if (len == 2 && m == 1) return nums[0] > nums[1] ? nums[0] : nums[1];*/ if (m == 0) { int res = 0; for (int i = 0; i < len; i++) res += nums[i]; return res; } long long result = 9999999999; for (int i = 1; i < len; i++) { vector<int> left; vector<int> right; for (int j = 0; j < i; j++) left.push_back(nums[j]); for (int k = i; k < len; k++) right.push_back(nums[k]); for (int leftm = 0; leftm <= m - 1; leftm++) { int rightm = m - 1 - leftm; int l = split(left, leftm); int r = split(right, rightm); int sum = l > r ? l : r; result = (result < sum) ? result : sum; } } return result; } int splitArray(vector<int>& nums, int m) { return split(nums, m - 1); }};
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