Leetcode 410. Split Array Largest Sum (Binary Search经典)

410. Split Array Largest Sum
     Hard
     8.9K
     184
     Companies
     Given an integer array nums and an integer k, split nums into k non-empty subarrays such that the largest sum of any subarray is minimized.

Return the minimized largest sum of the split.

A subarray is a contiguous part of the array.

Example 1:

Input: nums = [7,2,5,10,8], k = 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: nums = [1,2,3,4,5], k = 2
Output: 9
Explanation: There are four ways to split nums into two subarrays.
The best way is to split it into [1,2,3] and [4,5], where the largest sum among the two subarrays is only 9.

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] <= 106
1 <= k <= min(50, nums.length)

解法1：Binary Search。
不过感觉这题题目似乎有问题，应该不是刚好划分成k个组，而是<=k个组都可以。

class Solution {
public:
    int splitArray(vector<int>& nums, int k) {
        int n = nums.size();
        int totalSum = 0, maxNum = 0;
        for (auto num : nums) {
            maxNum = max(maxNum, num);
            totalSum += num;
        }
        int start = maxNum, end = totalSum;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (numSubArray(nums, mid) <= k) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if (numSubArray(nums, start) <= k) return start;
        return end;
    }
private:
    // return num of subarrays with maximum sum of each subarray
    int numSubArray(vector<int>& nums, int maxSum) {
        int n = nums.size();
        int res = 0, sum = 0;
        for (int i = 0; i < n; i++) {
            sum += nums[i];
            if (sum > maxSum) {
                sum = nums[i];
                res ++;
            }
        }
        if (sum > 0) res++;
        return res;
    }
};
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