410. Split Array Largest Sum

Given an array which consists of non-negative integers and an integer m, you can split the array intom non-empty continuous subarrays. Write an algorithm to minimize the largest sum among thesem 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 = 2Output:18Explanation: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.

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将给定的序列分成m个子序列，使得各个序列的总和的最大值最小，求出这个最小的最大值。看了discuss才知道怎样用二分法做，答案一定是在Max（序列的最大值）和sum（序列总和）之间，在这个范围内进行二分搜索。对于当前的值d，如果序列能分成m个和小于等于d的序列，则表示当前值是“有效的”，可以进一步减少来寻找最终答案；如果不能，即分成多于m个和小于等于d的序列，则当前值比答案小，增大之寻找最终答案。最后缩到一个值，判断这个值是否“有效”，“有效”的话答案是这个值，否则是这个值加1.

代码：

class Solution {public:int splitArray(vector<int>& nums, int m) {int sum = 0, Max = 0;for(auto num:nums){sum += num;Max = max(Max, num);}int l = Max, r = sum;while(l < r){int mid = l + (r - l) / 2;bool b = isvalid(nums, m, mid);if(b){r = mid - 1;}else{l = mid + 1;}}return isvalid(nums, m, l) ? l : l+1;}private:bool isvalid(vector<int>& nums, int m, int d){int sum = 0;for(auto num:nums){if(sum + num > d){--m;sum = 0;}if(m == 0) return false;sum += num;}return true;}};