【動態規劃】之最長公共子序列長度(難度:1星)

#include <stdio.h>

/**
 * 原題:
 * 給定兩個字符串S和T,求出這兩個字符串的最長公共子序列的長度
 * 例如X={A,B,C,B,D,A,B}，Y={B,D,C,A,B,A}
 * 則它們的lcs是4 ({B,C,B,A}和{B,D,A,B})。求出一個即可。
 */

#define MAX(x,y) (x>y?x:y)
#define SL 7
#define TL 6

static char s[SL+2] = " ABCBDAB";
static char t[TL+2] = " BDCABAB";

/**
 * 思路:
 * 設置子問題:求s的前i個字符組成的字符串與t的前j個字符組成的字符串的最大公共子序列長度
 * 找出邊界:顯然i或j等於0時,最大公共子序列長度爲0
 */
//解法1:遞歸
int solve_1(int i, int j){
    if (i == 0 || j == 0)
        return 0;
    if (s[i] == t[j])
        return MAX(solve_1(i-1, j-1) + 1, MAX(solve_1(i-1, j), solve_1(i, j-1)));
    return MAX(solve_1(i-1, j), solve_1(i, j-1));
}

//解法2:遞歸+記憶數組
static int memo[SL+1][TL+1];
int solve_2(int i, int j){
    if (i == 0 || j == 0)
        return memo[i][j] = 0;
    if (memo[i][j] > -1)
        return memo[i][j];
    if (s[i] == t[j])
        return memo[i][j] = MAX(solve_2(i-1, j-1) + 1, MAX(solve_2(i-1, j), solve_2(i, j-1)));
    return memo[i][j] = MAX(solve_2(i-1, j), solve_2(i, j-1));
}

//解法3:遞推
static int maxLen[SL+1][TL+1];
int solve_3(){

    for (int i = 1; i <= SL; ++i) {
        for (int j = 1; j <= TL; ++j) {
            if (s[i] == t[j])
                maxLen[i][j] = MAX(maxLen[i-1][j-1] + 1, MAX(maxLen[i-1][j], maxLen[i][j-1]));
            else
                maxLen[i][j] = MAX(maxLen[i-1][j], maxLen[i][j-1]);
        }
    }
    return maxLen[SL][TL];
}


int main() {
    printf("solve_1:%d\n", solve_1(SL, TL));

    for (int i = 0; i <= SL; ++i) {
        for (int j = 0; j <= TL ; ++j) {
            memo[i][j] = -1;
        }
    }
    printf("solve_2:%d\n", solve_2(SL, TL));
//    for (int i = 0; i <= SL; ++i) {
//        for (int j = 0; j <= TL ; ++j) {
//            printf("%d\t", memo[i][j]);
//        }
//        printf("\n");
//    }

    printf("solve_3:%d\n", solve_3());
//    for (int i = 0; i <= SL; ++i) {
//        for (int j = 0; j <= TL ; ++j) {
//            printf("%d\t", maxLen[i][j]);
//        }
//        printf("\n");
//    }

    return 0;
}

運行結果:

solve_1:4
solve_2:4
solve_3:4