leetcode题解-410. Split Array Largest Sum

题目：

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.

本题的目的是将数组分成m分，然后最小化各子数组的和的最大值。其实看到这个题目，我完全想不到他跟binary-search有什么关系。毕竟它的原理跟切分数组没有什么联系。然后就去看了别人的答案，才恍然大悟。原来binary-search还可以这么用==因为题目是要切分数组，然后最小化数组的和，所以呢，我们可以对数组的和进行binary-search。可以想到，数组和的最小值是最大元素的值，最大值是所有元素的和（注意溢出），然后我们对该区间进行二叉搜索，直到找到合适的结果。那么如何判断一个值是否满足呢，我们可以定义一个函数来判断数组是否可以满足该值。代码入下：

    public int splitArray(int[] nums, int m) {
        //取出最大值和最小值
        int max = 0; long sum = 0;
        for (int num : nums) {
            max = Math.max(num, max);
            sum += num;
        }
        if (m == 1) return (int)sum;
        //binary search
        long l = max; long r = sum;
        while (l <= r) {
            long mid = (l + r)/ 2;
            //调用valid函数判断当前值是否满足条件
            if (valid(mid, nums, m)) {
                r = mid - 1;
            } else {
                l = mid + 1;
            }
        }
        return (int)l;
    }
    public boolean valid(long target, int[] nums, int m) {
        int count = 1;
        long total = 0;
        //对数组进行分段求和，每当和大于target时，就重新求和，如果数组总数大于m，则说明target太小，需要增加，返回false
        for(int num : nums) {
            total += num;
            if (total > target) {
                total = num;
                count++;
                if (count > m) {
                    return false;
                }
            }
        }
        return true;
    }