DP.类似最大和。就是要记录正的和负的两种情况。
class Solution { public: int maxProduct(int A[], int n) { vector<int> positive(n); vector<int> minus(n); int result = A[0]; for (int i = 0; i < n; i++) { int minVal = A[i]; int maxVal = A[i]; if (i > 0) { minVal = min(minVal, positive[i - 1] * A[i]); minVal = min(minVal, minus[i - 1] * A[i]); maxVal = max(maxVal, positive[i - 1] * A[i]); maxVal = max(maxVal, minus[i - 1] * A[i]); } positive[i] = maxVal > 0 ? maxVal : 1; minus[i] = minVal < 0 ? minVal : 1; result = max(result, maxVal); } return result; } };