LeetCode 53/152 Maximum Subarray/Maximum Product Subarray---DP **
2015-04-09 16:57
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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
the contiguous subarray
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).
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
the contiguous subarray
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]);
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; } };
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