Leetcode 410. Split Array Largest Sum

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.

结果一定是在数组中最大的数和所有数字之和 的中间

注意： sum 用 long 来表示

之后使用二分法，用valid 函数 来检查是否成立

valid函数 从左到右 依次添加， total 表示 临时的和

如果total 大于 target， 把 total 设为 此刻的num，并 cnt++。如果cnt 大于 m，返回 false。（我们选的数小了，分的段数大于m，得把左边 mid +1）

反之， r = mid - 1

public int splitArray(int[] nums, int m) {
        int max = 0;
        long sum = 0;
        for (int num : nums) {
            max = Math.max(max, num);
            sum += num;
        }
        if (m == 1) return (int)sum;
        long l = max;
        long r = sum;
        while (l <= r) {
            long mid = (l + r) / 2;
            if (valid(mid, nums, m)) r = mid - 1;
            else l = mid + 1;
        }
        return (int)l;
    }
    
    private boolean valid(long target, int[] nums, int m) {
        int cnt = 1;
        long total = 0;
        for (int num : nums) {
            total += num;
            if (total > target) {
                total = num;
                cnt++;
                if (cnt > m) return false;
            }
        }
        return true;
    }