LeetCode Maximum Product Subarray
2014-12-27 14:03
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链接: https://oj.leetcode.com/problems/maximum-product-subarray/
求最大连续序列的乘积
因为偶数个负数相乘为正数,且任何数乘0等于0
.所以在计算的过程中需要维护两个数组,一个记录到当前位的最大乘积,一个记录到当前位的最小乘积..并初始化为A[i]
class Solution
{
public:
int maxProduct(int A[],int n)
{
int *maxdp=new int
;
int *mindp=new int
;
int ans=A[0];
for(int i=0;i<n;i++)
{
maxdp[i]=A[i];
mindp[i]=A[i];
}
for(int i=1;i<n;i++)
{
if(A[i]*mindp[i-1]>=A[i]*maxdp[i-1]&&A[i]*mindp[i-1]>=maxdp[i])
maxdp[i]=A[i]*mindp[i-1];
else if(A[i]*maxdp[i-1]>maxdp[i])
maxdp[i]=A[i]*maxdp[i-1];
if(A[i]*mindp[i-1]>=A[i]*maxdp[i-1]&&A[i]*maxdp[i-1]<=mindp[i])
mindp[i]=A[i]*maxdp[i-1];
else if(A[i]*mindp[i-1]<mindp[i])
mindp[i]=A[i]*mindp[i-1];
if(maxdp[i]>ans)
ans=maxdp[i];
}
return ans;
}
};
求最大连续序列的乘积
因为偶数个负数相乘为正数,且任何数乘0等于0
.所以在计算的过程中需要维护两个数组,一个记录到当前位的最大乘积,一个记录到当前位的最小乘积..并初始化为A[i]
class Solution
{
public:
int maxProduct(int A[],int n)
{
int *maxdp=new int
;
int *mindp=new int
;
int ans=A[0];
for(int i=0;i<n;i++)
{
maxdp[i]=A[i];
mindp[i]=A[i];
}
for(int i=1;i<n;i++)
{
if(A[i]*mindp[i-1]>=A[i]*maxdp[i-1]&&A[i]*mindp[i-1]>=maxdp[i])
maxdp[i]=A[i]*mindp[i-1];
else if(A[i]*maxdp[i-1]>maxdp[i])
maxdp[i]=A[i]*maxdp[i-1];
if(A[i]*mindp[i-1]>=A[i]*maxdp[i-1]&&A[i]*maxdp[i-1]<=mindp[i])
mindp[i]=A[i]*maxdp[i-1];
else if(A[i]*mindp[i-1]<mindp[i])
mindp[i]=A[i]*mindp[i-1];
if(maxdp[i]>ans)
ans=maxdp[i];
}
return ans;
}
};
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