学步园

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.

解题：

因为0比较影响结果，对0进行特判，凡是出现0的地方，重新计算。

每次计算的方式是计算le~ri之间的最大值，从左边往右记录le~x的积和从右往左扫记录x~ri的积，取个最大值就好了

代码：

class Solution {
public:
    int maxProduct(int A[], int n) {
        if(!n) return 0;
        int le = 0, ans = A[0];
        memset(sumL, 0, sizeof(int) * n);
        memset(sumR, 0, sizeof(int) * n);
        for(int i = 0; i < n; i ++) {
            if(!A[i]) {
                ans = max(ans, gao(le, i - 1, A));
                le = i + 1;
                ans = max(ans, 0);
            }
        }
        ans = max(ans, gao(le, n - 1, A));
        return ans;
    }
private:
    static const int SIZE = 100000;
    int sumL[SIZE], sumR[SIZE];
    int gao(int le, int ri, int A[]) {
        if(ri < le) return INT_MIN; //之前返回A[le]，但le可能超过边界n，导致出错
        sumL[le] = A[le], sumR[ri] = A[ri];
        for(int i = le + 1; i <= ri; i ++)
            sumL[i] = sumL[i - 1] * A[i];
        for(int i = ri - 1; i >= le; i --)
            sumR[i] = sumR[i + 1] * A[i];
        if(sumL[ri] > 0) return sumL[ri];
        int sum = A[le];
        for(int i = le; i <= ri; i ++) {
            if(A[i] <= 0) {
                if(i != le)
                    sum = max(sum, sumL[i - 1]);
                if(i != ri)
                    sum = max(sum, sumR[i + 1]);
            }
        }
        return sum;
    }
};