[LeetCode]Split Array Largest Sum

题目描述：

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:
Given m satisfies the following constraint: 1 ≤ m ≤ length(nums) ≤ 14,000.

Examples:

Input:nums = [1,2,3,4,5]m = 2

Output:
9

Explanation:
There are four ways to split nums into two subarrays.
The best way is to split it into [1,2,3] and [4,5],
where the largest sum among the two subarrays is only 9.

题目大意：

给定一个非负整数数组与一个整数m，你可以将数组拆分成m个非空连续子数组。编写算法使得这m个子数组各自的和的最大值最小化。

注意：

m满足如下约束：1 ≤ m ≤ length(nums) ≤ 14,000

解题思路：

二分枚举答案（Binary Search）

将数组nums拆分成m个子数组，每个子数组的和不小于sum(nums) / m，不大于sum(nums)

又因为数组nums中只包含非负整数，因此可以通过二分法在上下界内枚举答案。

时间复杂度O(n * log m)，其中n是数组nums的长度，m为数组nums的和

Python代码：

class Solution(object):
    def splitArray(self, nums, m):
        """
        :type nums: List[int]
        :type m: int
        :rtype: int
        """
        size = len(nums)
        high = sum(nums)
        low = high / m
        while low <= high:
            mid = (low + high) / 2
            n = m
            cnt = 0
            valid = True
            for x in range(size):
                if nums[x] > mid:
                    valid = False
                    break
                if cnt + nums[x] > mid:
                    n -= 1
                    cnt = 0
                cnt += nums[x]
                if n == 0:
                    valid = False
                    break
            if valid:
                high = mid - 1
            else:
                low = mid + 1
        return low