leetcode 152. Maximum Product Subarry

这道题求的是乘积的最大值的，那么依照之前的和的最大值类似的做法的，乘积的最大值可能是在最大值*当前值和最小值*当前值和当前值三者之间取得的最大值的，那么使用两个变量来保存每一步的最大最小值的。

class Solution {
public:int maxProduct(vector<int>& nums) {if(nums.size()<=0) return 0;int maxn,minn,res;maxn=minn=res=nums[0];for(int i=1;i<nums.size();i++){if(nums[i]>0){maxn=max(maxn*nums[i],nums[i]);minn=min(minn*nums[i],nums[i]);}else{int t=maxn; // 注意这里改变了最大值的maxn=max(minn*nums[i],nums[i]);minn=min(t*nums[i],nums[i]);}res=max(maxn,res);}return res;}
};

从上面看出还可以更简短代码一些的，当小于0的时候就可以直接进行交换最大和最小值的，也不需要利用中间变量来重新替换的。

class Solution {
public:int maxProduct(vector<int>& nums) {if(nums.size()<=0) return 0;int maxn,minn,res;maxn=minn=res=nums[0];for(int i=1;i<nums.size();i++){if(nums[i]<0) swap(maxn,minn);maxn=max(maxn*nums[i],nums[i]);minn=min(minn*nums[i],nums[i]);res=max(maxn,res);}return res;}
};

转载于:https://www.cnblogs.com/newnoobbird/p/9629053.html

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