【LeetCode】152. Maximum Product Subarray && 53. Maximum Subarray
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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
.
对于Product Subarray,要考虑到一种特殊情况,即负数和负数相乘:如果前面得到一个较小的负数,和后面一个较大的负数相乘,得到的反而是一个较大的数,如{2,-3,-7},所以,我们在处理乘法的时候,除了需要维护一个局部最大值,同时还要维护一个局部最小值,由此,可以写出如下的转移方程式:
max_copy[i] = max_local[i]
max_local[i + 1] = Max(Max(max_local[i] * A[i], A[i]), min_local * A[i])
min_local[i + 1] = Min(Min(max_copy[i] * A[i], A[i]), min_local * A[i])
class Solution {public: int maxProduct(vector<int>& nums) { int len = nums.size(); if(len==0) return 0; if(len==1) return nums[0]; int maxLocal = nums[0]; int minLocal = nums[0]; int maxAll = nums[0]; for(int i = 1; i < nums.size(); i++) { int max_copy = maxLocal; maxLocal = max(max(nums[i] * maxLocal, nums[i]), nums[i] * minLocal); minLocal = min(min(nums[i] * max_copy, nums[i]), nums[i] * minLocal); maxAll = max(maxAll, maxLocal); } return maxAll; }};
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
.
用动态规划的方法,就是要找到其转移方程式,也叫动态规划的递推式,动态规划的解法无非是维护两个变量,局部最优和全局最优,我们先来看Maximum SubArray的情况,如果遇到负数,相加之后的值肯定比原值小,但可能比当前值大,也可能小,所以,对于相加的情况,只要能够处理局部最大和全局最大之间的关系即可,对此,写出转移方程式如下:
local[i + 1] = Max(local[i] + A[i], A[i]);
global[i + 1] = Max(local[i + 1], global[i]);
class Solution {public: int maxSubArray(vector<int>& nums) { int len = nums.size(); if(len == 0) return 0; if(len==1) return nums[0]; int local = nums[0]; int all = nums[0]; for(int i=1; i < nums.size(); i++) { local =max((local+nums[i]),nums[i]); all = max(local, all); } return all; }};
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