[LeetCode]Maximum Product Subarray

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今天準備再刷一遍leetcode，發現出了一道新題。剛開始想貪心一遍O(n)實現，沒有想出來。於是從頭開始想，暴力雙迴圈，O(n*n)肯定沒有問題。然後下午坐車的時候又想到了可以用二分法，後來終於又想出來了O(n)的實現，如下所示，不過怎麼感覺自己的實現這麼麻煩呢，對0考慮的太多了。後面貼的別人的演算法就比較簡潔。

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.

int maxProduct(int A[], int n) {        int res = A[0];    int nmax = A[0] < 0 ? A[0] : 0;    int pmax = A[0] > 0 ? A[0] : 0;        for (int i = 1; i < n; i++)    {        if (A[i] == 0)        {            nmax = pmax = 0;        }        else if (A[i] > 0)        {            nmax *= A[i];            pmax = pmax != 0 ? pmax * A[i] : A[i];        }        else        {            swap(pmax, nmax);            nmax = nmax != 0 ? nmax * A[i] : A[i];            pmax *= A[i];        }        res = max(res, pmax);    }    return res;}

http://blog.csdn.net/sbitswc/article/details/39546719

public int maxProduct2(int[] A) {if(A.length<=0) return 0;if(A.length==1) return A[0];int curMax = A[0];int curMin = A[0];int ans = A[0];for(int i=1;i<A.length;i++){int temp = curMin *A[i];curMin = Math.min(A[i], Math.min(temp, curMax*A[i]));curMax = Math.max(A[i], Math.max(temp, curMax*A[i]));ans = Math.max(ans, curMax);}return ans;}

思路也很清晰：

Analysis:

similar like Maximum Subarray question

difference is the max value could be get from 3 situations

current maxValue * A[i]  if A[i]>0

current minValue * A[i]  if A[i]<0

A[i]  

We need to record current maxValue, current minValue and update them every time get the new product

[LeetCode]Maximum Product Subarray