LeetCode #410. Split Array Largest Sum

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本文介绍了一种算法，用于将数组分割成m个连续子数组，以最小化这些子数组的最大和。通过二分查找和贪心策略，该算法有效地解决了问题。示例中，数组[7,2,5,10,8]被分为[7,2,5]和[10,8]，最大和为18。

题目描述：

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.

class Solution {
public:
    int splitArray(vector<int>& nums, int m) {
        long long left=INT_MIN;
        long long right=0;
        for(int num:nums) 
        {
            left=max(left,(long long)num);
            right+=num;
        }
        // 思路是二分结果，根据最少可分的子数组数，确定搜索方向
        // 每个子数组和都要小于或等于mid，用贪心的方法求出最少需要分成几份
        while(left<right)
        {
            long long mid=(left+right)/2;
            int count=count_subarray(nums,mid);
            // 必须把数组分成大于m份，说明mid过小
            if(count>m) left=mid+1;
            else right=mid;
        }
        return left;
    }
    
    int count_subarray(vector<int>& nums, int target)
    {
        int count=1; // 注意count从1开始
        long long sum=0;
        for(int i=0;i<nums.size();i++)
        {
            if(sum+nums[i]>target)
            {
                sum=0;
                count++;
            }
            sum+=nums[i];
        }
        return count;
    }
};