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;
}