题目:Maximum Product Subarray
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.
这道题属于动态规划的题型,之前常见的是Maximum SubArray,现在是Product Subarray,不过思想是一致的。 当然不用动态规划,常规方法也是可以做的,但是时间复杂度过高(TimeOut),像下面这种形式:
1 // 思路:用两个指针来指向字数组的头尾
2 int maxProduct(int A[], int n)
3 {
4 assert(n > 0);
5 int subArrayProduct = -32768;
6
7 for (int i = 0; i != n; ++ i) {
8 int nTempProduct = 1;
9 for (int j = i; j != n; ++ j) {
10 if (j == i)
11 nTempProduct = A[i];
12 else
13 nTempProduct *= A[j];
14 if (nTempProduct >= subArrayProduct)
15 subArrayProduct = nTempProduct;
16 }
17 }
18 return subArrayProduct;
19 }
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]);
对应代码如下:
1 int maxSubArray(int A[], int n)
2 {
3 assert(n > 0);
4 if (n <= 0)
5 return 0;
6 int global = A[0];
7 int local = A[0];
8
9 for(int i = 1; i != n; ++ i) {
10 local = MAX(A[i], local + A[i]);
11 global = MAX(local, global);
12 }
13 return global;
14 }
而对于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])
对应代码如下:
1 #define MAX(x,y) ((x)>(y)?(x):(y))
2 #define MIN(x,y) ((x)<(y)?(x):(y))
3
4 int maxProduct1(int A[], int n)
5 {
6 assert(n > 0);
7 if (n <= 0)
8 return 0;
9
10 if (n == 1)
11 return A[0];
12 int max_local = A[0];
13 int min_local = A[0];
14
15 int global = A[0];
16 for (int i = 1; i != n; ++ i) {
17 int max_copy = max_local;
18 max_local = MAX(MAX(A[i] * max_local, A[i]), A[i] * min_local);
19 min_local = MIN(MIN(A[i] * max_copy, A[i]), A[i] * min_local);
20 global = MAX(global, max_local);
21 }
22 return global;
23 }
总结:动态规划题最核心的步骤就是要写出其状态转移方程,但是如何写出正确的方程式,需要我们不断的实践并总结才能达到。