410. Split Array Largest Sum（第十週）

Description：

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.

解題思路：

看到這道題第一想法是排序，讓數組裏的數字從小到大排列，然後再進行排列。然後看到題目意思是並不用排序，只用分塊就好。問題來了，怎麼分塊呢，遍歷找出所有的m個組合並不太實際。藉助了別人的思路，利用二分搜索來做這道題。據上述例子[7,2,5,10,8]，找出最大的一個數10，以及整個數組的和32，則解在[10,32]裏出現。取mid = 21，找到7,2,5 和 10,8。可以分則降低mid，取mid = 15，找到7,2,5和10和8，發現數組大於2，則升高mid，取mid = (15+21)/2 ……到最後直到left>=right 循環結束。返回的left值就是解，這是因爲right是保證了倒數第二次的數組能滿足的情況的值，當到最後一個數字，則需要left來滿足增大mid

#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    int splitArray(vector<int>& nums, int m) {
        int left = 0, right = 0;
        for(int i = 0; i < nums.size(); i++){
        	left   = max(left, nums[i]);
        	right += nums[i];
		}
		while(left < right){
			int mid = left + (right - left)/2;
			if(is_ok(nums, m, mid)) right = mid;
			else left = mid+1;
		}
		return left;
    }
    
    bool is_ok(vector<int>& nums, int m, int mid){
    	int array_count = 1;
    	int temp_sum = 0;
    	for(int i = 0; i < nums.size(); i++){
    		temp_sum += nums[i];
    		if(temp_sum > mid){
    			array_count++;
    			temp_sum = nums[i];
    			if(array_count > m) return false;
			}
		}
		return true;
	}
};