题目描述:

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:

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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.

这道题我是采用二分搜索的方法解的. 虽然要想到对整个int型范围进行二分搜索还是不太容易, 但是只要想到这一点, 问题就基本解决了, 剩下的就可以通过贪心来判断数组能不能被分成m个并且保证最大的和不超过一个给定的值. 二分搜索上界是INT_MAX, 而下界是数组中的最大值.

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class Solution {
public:
    int splitArray(vector<int>& nums, int m) {
        int len = nums.size();
        int maxElement = 0;
        for(auto i : nums){
            maxElement = max(maxElement, i);
        }
        long long left = maxElement, right = INT_MAX, mid = (left + right) / 2;
        while(left < right){
            if(canSplit(nums, mid, m)){
                right = mid;
            }
            else{
                left = mid + 1;
            }
            mid = (left + right) / 2;
        }
        return left;
    }
    
    bool canSplit(vector<int> &nums, long long target, int m){
        long long sum = 0;
        int splitArrs = 0;
        for(int i = 0; i < nums.size() && splitArrs <= m; i++){
            if(sum + nums[i] > target){
                sum = nums[i];
                splitArrs++;
            }
            else if(sum + nums[i] == target){
                sum = 0;
                splitArrs++;
            }
            else{
                sum += nums[i];
            }
        }
        if(sum > 0) splitArrs++;
        return splitArrs <= m;
    }
};