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.
题意：#
题目中给出一个（至少包含一个元素）整形数组，求一个子数组（元素连续），使得其元素之积最大。最直接了当的方法，当然是暴力穷举，但是其O(n^2)是无法通过 OJ 评判的。
C语言实现如下：
int maxProduct(int* nums, int numsSize) {
int nMaxPro = nums[0];
int max_tmp = nums[0];
int min_tmp = nums[0];
for(int i = 1; i< numsSize; ++i){
int a = nums[i] * max_tmp;
int b = nums[i] * min_tmp;
max_tmp = (( a > b ? a : b ) > nums[i]) ? ( a > b ? a : b ) : nums[i];
min_tmp = (( a < b ? a : b ) < nums[i]) ? ( a < b ? a : b ) : nums[i];
nMaxPro = (nMaxPro > max_tmp) ? nMaxPro : max_tmp;
}
return nMaxPro;
}C++实现如下：
class Solution {
public:
int maxProduct(vector<int>& nums) {
int nMaxPro = nums[0];
int max_tmp = nums[0];
int min_tmp = nums[0];
for(int i = 1; i< nums.size(); ++i){
int a = nums[i] * max_tmp;
int b = nums[i] * min_tmp;
max_tmp = (( a > b ? a : b ) > nums[i]) ? ( a > b ? a : b ) : nums[i];
min_tmp = (( a < b ? a : b ) < nums[i]) ? ( a < b ? a : b ) : nums[i];
nMaxPro = (nMaxPro > max_tmp) ? nMaxPro : max_tmp;
}
return nMaxPro;
}
};Java实现如下：
public class Solution {
public int maxProduct(int[] nums) {
int nMaxPro = nums[0];
int max_tmp = nums[0];
int min_tmp = nums[0];
for(int i = 1; i< nums.length; ++i){
int a = nums[i] * max_tmp;
int b = nums[i] * min_tmp;
max_tmp = (( a > b ? a : b ) > nums[i]) ? ( a > b ? a : b ) : nums[i];
min_tmp = (( a < b ? a : b ) < nums[i]) ? ( a < b ? a : b ) : nums[i];
nMaxPro = (nMaxPro > max_tmp) ? nMaxPro : max_tmp;
}
return nMaxPro;
}
}