leetcode--Maximum Product Subarray

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Find the contiguous subarray within an array (containing at least one number) which has the largest product.

For example, given the array [2,3,-2,4],
the contiguous subarray [2,3] has the largest product = 6.

题意：查询数组中有最大乘积的连续子数组。

分类：数组，动态规划

解法1：动态规划。由于乘积可能是负数，所以要保留两个数组，分别代表当前最大值和当前最小值。

递归规律：

max_arr[i] = Math.max(Math.max(max_arr[i-1]*nums[i],min_arr[i-1]*nums[i]),nums[i]);
min_arr[i] = Math.min(Math.min(max_arr[i-1]*nums[i],min_arr[i-1]*nums[i]),nums[i]);

public class Solution {    public int maxProduct(int[] nums) {        int len = nums.length;        int[] max_arr = new int[len];        int[] min_arr = new int[len];        int max = nums[0];        max_arr[0] = nums[0];        min_arr[0] = nums[0];        for(int i=1;i<len;i++){            max_arr[i] = Math.max(Math.max(max_arr[i-1]*nums[i],min_arr[i-1]*nums[i]),nums[i]);            min_arr[i] = Math.min(Math.min(max_arr[i-1]*nums[i],min_arr[i-1]*nums[i]),nums[i]);            max = Math.max(max,max_arr[i]);        }        return max;    }}

优化一下代码，因为每次只要保留当前的最大值和最小值即可

public class Solution {    public int maxProduct(int[] nums) {if(nums.length==0) return 0;if(nums.length==1) return nums[0];int curMax = nums[0];int curMin = nums[0];int res = nums[0];for(int i=1;i<nums.length;i++){int t = curMin*nums[i];curMin = Math.min(nums[i], Math.min(curMax*nums[i], t));curMax = Math.max(nums[i], Math.max(curMax*nums[i], t));res = Math.max(res,curMax);}return res;    }}

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