LeetCode 53/152 Maximum Subarray/Maximum Product Subarray---DP **

一：Maximum Subarray

题目：

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.

链接：https://leetcode.com/problems/maximum-subarray/

分析：这题就是求最大连续字串的和，或者说是求和最大的连续子数组

此题最简单的是暴力，但是暴力为O(N^2)， 超时。另外可以采用动态规划来求解，关键是如何转换为动态规划问题，即如何表述成动态规划问题。

我们用local[i]来表示以A[i]结尾的连续子数组的最大和，那么local[0] = A[0],,  local[1]等于要么local[0]+A[1]  要么就等于A[1]本身，关键看两者谁更大，于是便得到递归式local[i+1] = max(local[i]+A[i], A[i]),,,,local数组中的最大值即为所求。 迭代的过程中无需数组 因此时间复杂度为O(N), 空间复杂度为O(1).

class Solution {
public:
    int maxSubArray(int A[], int n) {
        if(n == 0) return 0;
        int sum = A[0];         // 以A[0]结束的连续子数组和最大值
        int maxResult = A[0];
        for(int i = 1; i < n; i++){
            sum = max(A[i],A[i]+sum);      // 以A[i]结束的连续子数组和最大值
            if(maxResult < sum) maxResult = sum; 
        }
        return maxResult;
    }

};

二：Maximum Product Subarray

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.

链接：https://leetcode.com/problems/maximum-product-subarray/

分析：此题我们不仅要保存一个最大值，还得保存一个最小值，因为负负相乘等于正数，比如-2 4 -5，即localMIn[i]表示以i结尾的连续字串乘积的最小值，而localMax[i]表示以i结尾的连续字串乘积的最大值，那么localMax[i] = max(max(localMin[i]*A[i], A[i]*localMax[i]), A[i]); 而localMin[i]=min(min(localMin[i]*A[i], A[i]*localMax[i]), A[i]);

class Solution {
public:
    int maxProduct(int A[], int n) {
        if(n == 0) return 0;
        int result = A[0];
        int minProduct = A[0];
        int maxProduct = A[0];
        for(int i = 1; i < n; i++){
            int temp = max(A[i]*maxProduct, A[i]*minProduct);
            temp = max(temp, A[i]);
            minProduct = min(A[i]*maxProduct, A[i]*minProduct);
            minProduct = min(minProduct, A[i]);
            maxProduct = temp;
            result = max(maxProduct, result);
        }
        return result;
        
    }
};