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]);

 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 == ) return ;
         if (nums.length == ) return nums[];

         int[] min = new int[nums.length];
         int[] max = new int[nums.length];

         min[] = nums[];
         max[] = nums[];

         for (int i = ; i < nums.length; i++) {
             min[i] = min(nums[i], min[i - ] * nums[i], max[i - ] * nums[i]);
             max[i] = max(nums[i], min[i - ] * nums[i], max[i - ] * nums[i]);
         }

         int tempMax = max[];
         for (int i = ; i < max.length; i++) {
             if (tempMax < max[i]) {
                 tempMax = max[i];
             }
         }
         return tempMax;

     }

     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));
     }
 }

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 == ) return ;
        if (nums.length == ) return nums[];

        int pre_temp_min = nums[];
        int pre_temp_max = nums[];
        int global_max = nums[];

        for (int i = ; 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));
    }
}

转载请注明出处：cnblogs.com/beiyeqingteng/