添加内容

This commit is contained in:
YDZ
2020-08-08 09:17:26 +08:00
parent 5015cc6b7b
commit fdba30f0c4
734 changed files with 4024 additions and 2129 deletions

View File

@ -0,0 +1,52 @@
# [152. Maximum Product Subarray](https://leetcode.com/problems/maximum-product-subarray/)
## 题目
Given an integer array `nums`, find the contiguous subarray within an array (containing at least one number) which has the largest product.
**Example 1**:
Input: [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.
**Example 2**:
Input: [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.
## 题目大意
给定一个整数数组 nums ,找出一个序列中乘积最大的连续子序列(该序列至少包含一个数)。
## 解题思路
- 给出一个数组,要求找出这个数组中连续元素乘积最大的值。
- 这一题是 DP 的题,状态转移方程是:最大值是 `Max(f(n)) = Max( Max(f(n-1)) * n, Min(f(n-1)) * n)`;最小值是 `Min(f(n)) = Min( Max(f(n-1)) * n, Min(f(n-1)) * n)`。只要动态维护这两个值,如果最后一个数是负数,最大值就在负数 * 最小值中产生,如果最后一个数是正数,最大值就在正数 * 最大值中产生。
## 代码
```go
package leetcode
func maxProduct(nums []int) int {
minimum, maximum, res := nums[0], nums[0], nums[0]
for i := 1; i < len(nums); i++ {
if nums[i] < 0 {
maximum, minimum = minimum, maximum
}
maximum = max(nums[i], maximum*nums[i])
minimum = min(nums[i], minimum*nums[i])
res = max(res, maximum)
}
return res
}
```