【LeetCode】【Solution】Maximum Product Subarray

标签：动态规划   最大连续乘积   leetcode   algorithm   

【题目】

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.

参考 http://segmentfault.com/blog/riodream/1190000000691766 

public class Solution {
    public int maxProduct(int[] A) {
        if (A.length == 0) return 0;
        if (A.length == 1) return A[0];
        
        int max_ending_here = 0;
        int min_ending_here = 0;
        int max_so_far = 0;
        
        for (int i = 0; i < A.length; i++) {
            if (A[i] > 0) {
                max_ending_here = Math.max(max_ending_here * A[i], A[i]);
                min_ending_here = Math.min(min_ending_here * A[i], A[i]);
            } else if (A[i] == 0) {
                max_ending_here = 0;
                min_ending_here = 0;
            } else {
                int temp = max_ending_here;
                max_ending_here = Math.max(min_ending_here*A[i], A[i]);
                min_ending_here = Math.min(temp * A[i], A[i]);
            }
            
            if (max_ending_here > max_so_far) {
                max_so_far = max_ending_here;
            }
        }
        
        return max_so_far;
    }
}

Let us denote that:

f(k) = Largest product subarray, from index 0 up to k.

Similarly,

g(k) = Smallest product subarray, from index 0 up to k.

Then,

f(k) = max( f(k-1) * A[k], A[k], g(k-1) * A[k] )
g(k) = min( g(k-1) * A[k], A[k], f(k-1) * A[k] )

public int maxProduct(int[] A) {
   assert A.length > 0;
   int max = A[0], min = A[0], maxAns = A[0];
   for (int i = 1; i < A.length; i++) {
      int mx = max, mn = min;
      max = Math.max(Math.max(A[i], mx * A[i]), mn * A[i]);
      min = Math.min(Math.min(A[i], mx * A[i]), mn * A[i]);
      maxAns = Math.max(max, maxAns);
   }
   return maxAns;
}

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【LeetCode】【Solution】Maximum Product Subarray

标签：动态规划   最大连续乘积   leetcode   algorithm   

原文地址：http://blog.csdn.net/ljiabin/article/details/41013619