[Swift]LeetCode410. 分割数组的最大值 | Split Array Largest Sum

原文地址：https://www.cnblogs.com/strengthen/p/10330269.html 

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.

---

给定一个非负整数数组和一个整数 m，你需要将这个数组分成 m 个非空的连续子数组。设计一个算法使得这 m 个子数组各自和的最大值最小。

注意:
数组长度 n 满足以下条件:

• 1 ≤ n ≤ 1000
• 1 ≤ m ≤ min(50, n)

示例:

输入:
nums = [7,2,5,10,8]
m = 2

输出:
18

解释:
一共有四种方法将nums分割为2个子数组。
其中最好的方式是将其分为[7,2,5] 和 [10,8]，
因为此时这两个子数组各自的和的最大值为18，在所有情况中最小。

---

12ms

class Solution {
    func splitArray(_ nums: [Int], _ m: Int) -> Int {
        var nums = nums
        var left:Int = 0
        var right:Int = 0
        for i in 0..<nums.count
        {
            left = max(left, nums[i])
            right += nums[i]
        }
        while (left < right)
        {
            var mid:Int = left + (right - left) / 2
            if canSplit(&nums, m, mid)
            {
                right = mid
            }
            else
            {
                left = mid + 1
            }
        }
        return left
    }
    
    func canSplit(_ nums:inout [Int], _ m: Int, _ sum: Int) -> Bool
    {
        var cnt:Int = 1
        var curSum:Int = 0
        for i in 0..<nums.count
        {
            curSum += nums[i]
            if curSum > sum
            {
                curSum = nums[i]
                cnt += 1
                if cnt > m
                {
                    return false
                }
            }
        }
        return true
    }
}