LeetCode: Maximum Product Subarray && Maximum Subarray

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Maximum Product Subarray

Title:

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.

对于Product Subarray，要考虑到一种特殊情况，即负数和负数相乘：如果前面得到一个较小的负数，和后面一个较大的负数相乘，得到的反而是一个较大的数，如{2，-3，-7}，所以，我们在处理乘法的时候，除了需要维护一个局部最大值，同时还要维护一个局部最小值，由此，可以写出如下的转移方程式：

max_copy[i] = max_local[i]
max_local[i + 1] = Max(Max(max_local[i] * A[i], A[i]),  min_local * A[i])

min_local[i + 1] = Min(Min(max_copy[i] * A[i], A[i]),  min_local * A[i])

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int pmin = nums[0];
        int pmax = nums[0];
        int result = nums[0];
        for (int i = 1; i < nums.size(); i++){
            int tmax = pmax * nums[i];
            int tmin = pmin * nums[i];
            pmax = max(nums[i],max(tmax,tmin));
            pmin = min(nums[i],min(tmax,tmin));
            result = max(result,pmax);
        }
        return result;
    }
};

Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

class Solution{
public:
    int maxSubArray(int A[], int n) {
        int maxSum = A[0];
        int sum = A[0];
        for (int i = 1; i < n; i++){
            if (sum < 0)
                sum = 0;
            sum += A[i];
            maxSum = max(sum,maxSum);
        }
        return maxSum;
    }
};

LeetCode: Maximum Product Subarray && Maximum Subarray

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原文地址：http://www.cnblogs.com/yxzfscg/p/4537970.html