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92 lines
2.3 KiB
Markdown
Executable File
92 lines
2.3 KiB
Markdown
Executable File
# [410. Split Array Largest Sum](https://leetcode.com/problems/split-array-largest-sum/)
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## 题目
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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.
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**Note**:If n is the length of array, assume the following constraints are satisfied:
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- 1 ≤ n ≤ 1000
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- 1 ≤ m ≤ min(50, n)
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**Examples**:
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Input:
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nums = [7,2,5,10,8]
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m = 2
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Output:
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18
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Explanation:
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There are four ways to split nums into two subarrays.
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The best way is to split it into [7,2,5] and [10,8],
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where the largest sum among the two subarrays is only 18.
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## 题目大意
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给定一个非负整数数组和一个整数 m，你需要将这个数组分成 m 个非空的连续子数组。设计一个算法使得这 m 个子数组各自和的最大值最小。
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注意:
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数组长度 n 满足以下条件:
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- 1 ≤ n ≤ 1000
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- 1 ≤ m ≤ min(50, n)
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## 解题思路
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- 给出一个数组和分割的个数 M。要求把数组分成 M 个子数组，输出子数组和的最大值。
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- 这一题可以用动态规划 DP 解答，也可以用二分搜索来解答。这一题是二分搜索里面的 max-min 最大最小值问题。题目可以转化为在 `M` 次划分中，求一个 `x`，使得 `x` 满足：对任意的`S(i)`，都满足 `S(i) ≤ x`。这个条件保证了 `x` 是所有 `S(i)` 中的最大值。要求的是满足该条件的最小的 `x`。`x` 的搜索范围在 `[max, sum]` 中。逐步二分逼近 low 值，直到找到能满足条件的 low 的最小值，即为最终答案。
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## 代码
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```go
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package leetcode
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func splitArray(nums []int, m int) int {
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maxNum, sum := 0, 0
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for _, num := range nums {
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sum += num
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if num > maxNum {
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maxNum = num
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}
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}
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if m == 1 {
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return sum
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}
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low, high := maxNum, sum
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for low < high {
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mid := low + (high-low)>>1
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if calSum(mid, m, nums) {
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high = mid
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} else {
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low = mid + 1
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}
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}
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return low
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}
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func calSum(mid, m int, nums []int) bool {
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sum, count := 0, 0
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for _, v := range nums {
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sum += v
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if sum > mid {
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sum = v
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count++
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// 分成 m 块，只需要插桩 m -1 个
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if count > m-1 {
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return false
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}
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}
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}
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return true
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}
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``` |