152. Maximum Product Subarray
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题意: 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.
思路:这题和121. Best Time to Buy and Sell Stock后面提到的卡登算法有一些相似,这题的关键在于要保存最小数,因为最小数就是潜在的最大数,因为当最小数为负数乘以一个负数的时候就成了最大数,所以这题需要保存在小数和最大数。
代码:
class Solution {public: int maxProduct(vector<int>& nums) { if (nums.empty()) return 0; int p_max=nums[0],p_min=nums[0],ans=nums[0]; for (int i=1;i<nums.size();i++){ if (nums[i]<0) swap(p_min,p_max); p_max = max(nums[i],p_max*nums[i]); p_min = min(nums[i],p_min*nums[i]); ans = p_max>ans?p_max:ans; } return ans; }};
其中的那就swap很重要,免去了每次将最小*nums[i],最大*nums[i]和nums[i]一起比较。
在网上还看到一个很神奇的做法,特别记录下来:
class Solution { // author : s2003zy // weibo : http://weibo.com/574433433 // blog : http://s2003zy.com // Time : O(n) // Space : O(1) public: int maxProduct(int A[], int n) { int frontProduct = 1; int backProduct = 1; int ans = INT_MIN; for (int i = 0; i < n; ++i) { frontProduct *= A[i]; backProduct *= A[n - i - 1]; ans = max(ans,max(frontProduct,backProduct)); frontProduct = frontProduct == 0 ? 1 : frontProduct; backProduct = backProduct == 0 ? 1 : backProduct; } return ans; }};
这段代码没有错!它的巧妙之处在于分别从头和从尾部开始计算,对于数组中只有一个负数的情况,就是看负数两边谁的积更大,对于两个负数的情况,当然是总积最大,所以只需要考虑奇数个负数时,比如:’aBcDe’中(小写字母代表负数,大写代表正数),’aBcD’和’BcDe’谁更大,这就是代码从两头算起的含义,最后只要把有0的情况单独拉出来就好了。
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- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
- 152. Maximum Product Subarray
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