leetcode(152). Maximum Product Subarray
2017-08-08 13:32
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problem
Find the contiguous subarray within an array (containing at least onenumber) which has the largest product.
For example, given the array [2,3,-2,4], the contiguous subarray [2,3]
has the largest product = 6.
solution
O(n2)解法:class Solution(object): def maxProduct(self, nums): """ :type nums: List[int] :rtype: int """ ans = float('-inf') tmp = [1] for num in nums: for i, n in enumerate(tmp): tmp[i] = n*num ans = max(ans, tmp[i]) tmp.append(1) return ans
dp
这个问题也可以使用动态规划来求解,因为这个问题求的是乘法,因此就可以把整数分成负数,零和正数,子问题f(i)和g(i)定义为以nums[i]为结尾的子数组的乘积正最大值和负最小值,状态转移:如果nums[i+1]为零,那么f(i+1)和g(i+1)都等于零,
如果f(i), g(i) != 0:
nums[i+1] > 0,f(i+1) = f(i)*nums[i+1],g(i+1)=g(i)*nums[i+1]
nums[i+1] < 0,f(i+1) = g(i) * nums[i+1], g(i+1) = f(i) * nums[i+1]
具体逻辑如下:
class Solution(object): def maxProduct(self, nums): """ :type nums: List[int] :rtype: int """ if len(nums) == 1: return nums[0] if nums[0] >= 0: maxPos = [nums[0]] minNeg = [0] else: maxPos = [0] minNeg = [nums[0]] for i, num in enumerate(nums[1:], 1): if num >= 0: if maxPos[i-1] != 0: maxPos.append(maxPos[i-1]*num) else: maxPos.append(num) minNeg.append(minNeg[i-1]*num) else: if maxPos[i-1] != 0: minNeg.append(maxPos[i-1]*num) else: minNeg.append(num) maxPos.append(minNeg[i-1]*num) #不必担心0不是正确解,因为超过两个数时或者0是正确解或者有大于零的数 return max(maxPos)
超过了97%的提交,由于此题的dp中前一个子问题的结果,所以可以把它转化为循环形式,使用两个变量保存前一个子问题的解,每次循环时更新最大积,这样空间复杂度为O(1),代码没有相差太多,不再赘述。
总结
感觉子数组和子序列求最值问题都可以使用动态规划,但是对于找某个值的问题通常都是使用hash表,可能因为最值问题可以形成递推关系,而具体值不好递推。相关文章推荐
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