标签：within nbsp one turn ges 速度 blog sub res
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.
找出最大的相乘的数字，很简单，代码还可以优化的地方很多。但是速度还可以。
public class Solution { public int maxProduct(int[] nums) { int len = nums.length; if( nums.length == 1) return nums[0]; int[] result = new int[2]; int target = 0; int left_right = 0,right_left = 0; for( int i = 0 ; i < len ; i ++){ if( nums[i] == 0){ target = Math.max(target,result[0]); result[0] = 0; result[1] = 0; target = 0; }else if( nums[i] > 0 ){ if( result[0] != 0) result[0] *= nums[i]; else result[0] = nums[i]; }else if( nums[i] < 0 ){ if( result[1] == 0 ){ result[1] = nums[i]*(result[0] == 0?1:result[0]); result[0] = 0; } else{ if( result[0] == 0){ result[0] = result[1]*nums[i]; result[1] = 0; }else{ result[0] = result[0]*result[1]*nums[i]; result[1] = 0; } } } left_right = Math.max(Math.max(result[0],target),left_right); } target = 0; result[0] = 0; result[1] = 0; for( int i = len-1;i>=0;i--){ if( nums[i] == 0){ target = Math.max(target,result[0]); result[0] = 0; result[1] = 0; target = 0; }else if( nums[i] > 0 ){ if( result[0] != 0) result[0] *= nums[i]; else result[0] = nums[i]; }else if( nums[i] < 0 ){ if( result[1] == 0 ){ result[1] = nums[i]*(result[0] == 0?1:result[0]); result[0] = 0; } else{ if( result[0] == 0){ result[0] = result[1]*nums[i]; result[1] = 0; }else{ result[0] = result[0]*result[1]*nums[i]; result[1] = 0; } } } right_left = Math.max(Math.max(result[0],target),right_left); } return Math.max(left_right,right_left); } }
leetcode 152. Maximum Product Subarray --------- java
标签：within nbsp one turn ges 速度 blog sub res
原文地址：http://www.cnblogs.com/xiaoba1203/p/6079446.html
