LeetCode每日一题(410. Split Array Largest Sum)

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本文介绍了一种使用二分搜索算法解决将数组分割为m个连续子数组，并使这些子数组中的最大和最小化的问题。通过确定搜索范围，即数组总和与数组中最大元素之间的值，可以有效地找到最优解。

Given an array nums 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.

Example 1:

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.

Example 2:

Input: nums = [1,2,3,4,5], m = 2
Output: 9

Example 3:

Input: nums = [1,4,4], m = 3
Output: 4

Constraints:

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

首先我们能想到，最终答案的最大值就是整个nums的sum, 最小值则是sum / m和nums中的最大的元素相比较较大的那个。我们既然知道了取值范围，自然而然的就会想到用 binary search 的方法来查找这个值。剩下的看代码吧

impl Solution {
    pub fn split_array(nums: Vec<i32>, m: i32) -> i32 {
        let mut sum = 0;
        let mut maximum = 0;
        for i in 0..nums.len() {
            sum += nums[i];
            maximum = maximum.max(nums[i]);
        }
        let mut min = (sum / m).max(maximum);
        let mut max = sum;
        while min < max {
            let mid = (min + max) / 2;
            let mut count = 0;
            let mut curr = 0;
            for i in 0..nums.len() {
                if curr + nums[i] > mid {
                    count += 1;
                    curr = nums[i];
                } else {
                    curr += nums[i];
                }
            }
            count += 1;
            if count > m {
                min = mid + 1;
            } else {
                max = mid;
            }
        }
        min
    }
}