Leetcode | Maximum Product Subarray

Leetcode 加新题了

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.

整理思路不难，就是维护一个包含当前位置的最大正数和最小负数。在处理负数的时候需要注意。

class Solution {
public:
    int maxProduct(int arr[], int n) {
        if (n <= 0) return 0;
        int maxPos = 1, minNeg = 1, max = INT_MIN;
        for (int i = 0; i < n; ++i) {
            if (arr[i] > 0) {
                maxPos = maxPos * arr[i];
                minNeg = minNeg > 0 ? 1: minNeg * arr[i];
                if (maxPos > max) max = maxPos;
            } else if (arr[i] == 0) {
                maxPos = minNeg = 1;
                if (max < 0) max = 0;
            } else { // arr[i] < 0
                int tmp = maxPos;
                if (minNeg > 0) {
                    maxPos = 1;
                } else {
                    maxPos = minNeg * arr[i];
                    if (maxPos > max) max = maxPos;
                }
                minNeg = tmp * arr[i];
                if (minNeg > max) max = minNeg;
            }
        }
        return max;
    }
};

 看了一下solution之后，还有更简洁的写法。我还是好弱。每天都能学到新东西啊。

class Solution {
public:
    int maxProduct(int arr[], int n) {
        if (n <= 0) return 0;
        int maxSoFar = arr[0], minSoFar = arr[0], ans = arr[0];
        for (int i = 1; i < n; ++i) {
            int tmp = maxSoFar;
            maxSoFar = max(max(arr[i], maxSoFar * arr[i]), minSoFar * arr[i]);
            minSoFar = min(min(arr[i], tmp * arr[i]), minSoFar * arr[i]);
            if (maxSoFar > ans) ans = maxSoFar;
        }
        return ans;
    }
};