算法: 动态规划解最长公共子串

继上篇文章“算法：动态规划解 0-1 背包问题”，再发一篇

上篇： http://blog.youkuaiyun.com/span76/article/details/8033953

#include <iostream>
using namespace std;

char a[]="abecd";
char b[]= "abcd";

char result[512];
int rIdx = -1;
int max2 = 0;

int find(char a[], char b[], int aIdx, int bIdx) {
    if ( aIdx < 0 || bIdx < 0) {
        int tIdx = rIdx;
        if( max2 < tIdx+1  ){
            max2 = tIdx+1;
            while (tIdx>=0)
                cout << result[tIdx--];  // 打印最长的串, 当然如果后来发现更长的也会打印
            cout<< endl;
        }
        return 0;
    }

    if (  a[aIdx] == b[bIdx] ) {
        result[++rIdx] = a[aIdx];  // 为了打印
        int ret = 1 + find(a,b, aIdx-1, bIdx-1);
        result[--rIdx]; // 递归的魅力在于能复原现场, 如果匹配,没有找到的最长串, 现场还是会被恢复
        return ret;
    }
    else { // 对比下面的两种情况 
        int m = find(a,b, aIdx-1, bIdx);  
        int n = find(a,b, aIdx, bIdx-1);
        return m>n? m: n;  // 取最优,体现动态规划的特点
    }
}



int main() {
    cout << "!!!Max Common String!!!" << endl;

    int aLen = strlen(a);
    int bLen = strlen(b);
    cout << "1:" << a << endl;
    cout << "2:" << b << endl;

    int maxLen = find(a,b,aLen-1, bLen-1 );
    cout << endl << "max len is " << maxLen ;

    return 0;
}