152. Maximum Product Subarray

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

Example

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

分析：

因为这题要求乘积最大，而负数和负数相乘可是是正数，所以，我们必须得保存两个变量，即当到达A[i]时，前一个数A[i - 1]所对应的最大值和最小值max[i - 1] 和 min[i - 1]. max[i] 和min[i]和这些值有如下关系：

min[i] = min(nums[i], min[i - 1] * nums[i], max[i - 1] * nums[i]);
max[i] = max(nums[i], min[i - 1] * nums[i], max[i - 1] * nums[i]);

 1 public class Solution {
 2     /**
 3      * @param nums: an array of integers
 4      * @return: an integer
 5      * cnblogs.com/beiyeqingteng/
 6      */
 7     public int maxProduct(int[] nums) {
 8         if (nums == null || nums.length == 0) return 0;
 9         if (nums.length == 1) return nums[0];
10         
11         int[] min = new int[nums.length];
12         int[] max = new int[nums.length];
13         
14         min[0] = nums[0];
15         max[0] = nums[0];
16         
17         for (int i = 1; i < nums.length; i++) {
18             min[i] = min(nums[i], min[i - 1] * nums[i], max[i - 1] * nums[i]);
19             max[i] = max(nums[i], min[i - 1] * nums[i], max[i - 1] * nums[i]);
20         }
21         
22         int tempMax = max[0];
23         for (int i = 1; i < max.length; i++) {
24             if (tempMax < max[i]) {
25                 tempMax = max[i];
26             }   
27         }
28         return tempMax;
29         
30     }
31     
32     public int max(int a, int b, int c) {
33         return Math.max(a, Math.max(b, c));
34     }
35     
36     public int min(int a, int b, int c) {
37         return Math.min(a, Math.min(b, c));
38     }
39 }

更简洁的做法：

public class Solution {
    /**
     * @param nums: an array of integers
     * @return: an integer
     * cnblogs.com/beiyeqingteng/
     */
    public int maxProduct(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        
        int pre_temp_min = nums[0];
        int pre_temp_max = nums[0];
        int global_max = nums[0];
        
        for (int i = 1; i < nums.length; i++) {
            int temp_min = min(nums[i], pre_temp_min * nums[i], pre_temp_max * nums[i]);
            int temp_max = max(nums[i], pre_temp_min * nums[i], pre_temp_max * nums[i]);
            pre_temp_min = temp_min;
            pre_temp_max = temp_max;
            global_max = Math.max(global_max, temp_max);
        }
        return global_max;
    }
    
    public int max(int a, int b, int c) {
        return Math.max(a, Math.max(b, c));
    }
    
    public int min(int a, int b, int c) {
        return Math.min(a, Math.min(b, c));
    }
}

 

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posted @ 2016-07-01 05:44  北叶青藤  阅读(195)  评论(0)  编辑  收藏  举报