LintCode 1251：Split Array Largest Sum (Binary Search好题)

1251 · Split Array Largest Sum
Algorithms
Hard

Description
Given an array which consists of non-negative integers and an integer m, we are going to split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays.

If n is the length of array, assume the following constraints are satisfied:

1 ≤ n ≤ 1000
1 ≤ m ≤ min(50, n)
https://kns.cnki.net/kns8/defaultresult/index

Example
Example 1:

Input：[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.
Example 2:

Input：[1,4,4], m = 3
Output：4
Explanation：
There is a way to split nums into three subarrays.
The best way is to split it into [1], [4] and [4],
where the largest sum among the three subarrays is only 4.

解法1：binary search。
二分思想在下面的链接中解释得比较清楚。
https://www.cnblogs.com/grandyang/p/5933787.html
一些需要注意的地方：

1. can_split(nums, m, thresh) 表示nums[]数组能否分成m个连续子数组，使得其中最大连续子数组的和不超过thresh。如果nums[]数组已经能分成<m个连续子数组且其中最大连续子数组的和不超过thresh，那么nums[]数组肯定也能分成m个连续子数组且其中最大连续子数组的和不超过thresh。这就是为什么can_split() 函数里面最后当num_subarrs <= m时就直接返回true了。
2. 如果can_split(nums, m, thresh)成立，那么比thresh更大的thresh’肯定也成立，我们需要找到更小的thresh，所以取end=mid。
   最后我们先判断start，再判断end，因为我们要取小的那个。

class Solution {
public:
    /**
     * @param nums: a list of integers
     * @param m: an integer
     * @return: return a integer
     */
    int splitArray(vector<int> &nums, int m) {
        int n = nums.size();
        if (n == 0) return 0;
        int max_elem = *max_element(nums.begin(), nums.end());
        int sum = 0;
        for (int i = 0; i < n; ++i) sum += nums[i];
        int start = max_elem, end = sum;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (can_split(nums, m, mid)) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if (can_split(nums, m, start)) return start;
        if (can_split(nums, m, end)) return end;
    }

private:
    bool can_split(vector<int> &nums, int m, int thresh) {
        int n = nums.size();
        int index = 0, sum = 0;
        int num_subarrs = 1;
        
        while (index < n) {
            sum += nums[index];
            if (sum > thresh) {
                sum = nums[index];
                num_subarrs++; 
                if (num_subarrs > m) return false;
            }
            index++;
        }
        
     //   if (num_subarrs == m) return true;   //错误
     //   return false;                        //错误
        return true;
    }

};

解法2：DP。TBD。