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## 题目地址 | ||
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https://leetcode.com/problems/maximum-product-subarray/description/ | ||
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## 题目描述 | ||
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``` | ||
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. | ||
``` | ||
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## 思路 | ||
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> 这道题目的通过率非常低 | ||
这道题目要我们求解连续的 n 个数中乘积最大的积是多少。这里提到了连续,笔者首先 | ||
想到的就是滑动窗口,但是这里比较特殊,我们不能仅仅维护一个最大值,因此最小值(比如-20)乘以一个比较小的数(比如-10) | ||
可能就会很大。 因此这种思路并不方便。 | ||
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首先来暴力求解,我们使用两层循环来枚举所有可能项,这种解法的时间复杂度是O(n^2), 代码如下: | ||
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```js | ||
var maxProduct = function(nums) { | ||
let max = nums[0]; | ||
let temp = null; | ||
for (let i = 0; i < nums.length; i++) { | ||
temp = nums[i]; | ||
max = Math.max(temp, max); | ||
for (let j = i + 1; j < nums.length; j++) { | ||
temp *= nums[j]; | ||
max = Math.max(temp, max); | ||
} | ||
} | ||
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return max; | ||
}; | ||
``` | ||
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因此我们需要同时记录乘积最大值和乘积最小值,然后比较元素和这两个的乘积,去不断更新最大值。 | ||
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![152.maximum-product-subarray](../assets/problems/152.maximum-product-subarray.png) | ||
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这种思路的解法由于只需要遍历依次,其时间复杂度是O(n),代码见下方代码区。 | ||
## 关键点 | ||
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- 同时记录乘积最大值和乘积最小值 | ||
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## 代码 | ||
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```js | ||
/* | ||
* @lc app=leetcode id=152 lang=javascript | ||
* | ||
* [152] Maximum Product Subarray | ||
* | ||
* https://leetcode.com/problems/maximum-product-subarray/description/ | ||
* | ||
* algorithms | ||
* Medium (28.61%) | ||
* Total Accepted: 202.8K | ||
* Total Submissions: 700K | ||
* Testcase Example: '[2,3,-2,4]' | ||
* | ||
* 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. | ||
* | ||
*/ | ||
/** | ||
* @param {number[]} nums | ||
* @return {number} | ||
*/ | ||
var maxProduct = function(nums) { | ||
let max = nums[0]; | ||
let min = nums[0]; | ||
let res = nums[0]; | ||
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for (let i = 1; i < nums.length; i++) { | ||
let tmp = min; | ||
min = Math.min(nums[i], Math.min(max * nums[i], min * nums[i])); // 取最小 | ||
max = Math.max(nums[i], Math.max(max * nums[i], tmp * nums[i])); /// 取最大 | ||
res = Math.max(res, max); | ||
} | ||
return res; | ||
}; | ||
``` |
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