【LeetCode】152. Maximum Product Subarray && 53. Maximum Subarray

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 len = nums.size();
if(len==0) return 0;
if(len==1) return nums[0];
int maxLocal = nums[0];
int minLocal = nums[0];
int maxAll = nums[0];

for(int i = 1; i < nums.size(); i++)
{
int max_copy = maxLocal;
maxLocal = max(max(nums[i] * maxLocal, nums[i]), nums[i] * minLocal);
minLocal = min(min(nums[i] * max_copy, nums[i]), nums[i] * minLocal);
maxAll = max(maxAll, maxLocal);
}
return maxAll;
}
};

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

.

用动态规划的方法，就是要找到其转移方程式，也叫动态规划的递推式，动态规划的解法无非是维护两个变量，局部最优和全局最优，我们先来看Maximum SubArray的情况，如果遇到负数，相加之后的值肯定比原值小，但可能比当前值大，也可能小，所以，对于相加的情况，只要能够处理局部最大和全局最大之间的关系即可，对此，写出转移方程式如下：

local[i + 1] = Max(local[i] + A[i], A[i]);

global[i + 1] = Max(local[i + 1], global[i]);

class Solution {
public:
int maxSubArray(vector<int>& nums) {
int len = nums.size();
if(len == 0) return 0;
if(len==1) return nums[0];
int local = nums[0];
int all = nums[0];
for(int i=1; i < nums.size(); i++)
{
local  =max((local+nums[i]),nums[i]);
all = max(local, all);
}
return all;
}
};