洛谷UVA536 二叉树重建 Tree Recovery

最新推荐文章于 2023-02-05 18:52:10 发布

原创最新推荐文章于 2023-02-05 18:52:10 发布 · 195 阅读

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文章标签：

#Python #C语言 #二叉树遍历 #输入格式 #字符串截取

https://www.luogu.com.cn/problem/UVA536

C语言：

字符串截取：strncpy(s1, s2 + i, j)：从s2的第 i+1 位开始截取 j 位，赋给s1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void PostOrder(char s1[100], char s2[100], int len)
{
    int i;
    char s11[100], s12[100], s21[100], s22[100];

    if (len==1)
    {
        printf("%c", s1[0]);
        return;
    }
    else if (len==0)
        return;

    for (i=0; i<len; i++)
    {
        if (s1[0] == s2[i])
        {
            if (i != 0)
            {
                strncpy(s11, s1 + 1, i);
                strncpy(s21, s2, i);
                s11[i] = '\0';
                s21[i] = '\0';
            }
            if (i != len - 1)
            {
                strncpy(s12, s1 + i + 1, len - i - 1);
                strncpy(s22, s2 + i + 1, len - i - 1);
                s12[len - i - 1] = '\0';
                s22[len - i - 1] = '\0';
            }
            //printf("%s %s %s %s\n", s11, s12, s21, s22);
            PostOrder(s11, s21, i);
            PostOrder(s12, s22, len - i - 1);
            break;
        }
    }
    printf("%c",s1[0]);
}


int main()
{
    int i;
    char s1[100], s2[100];

    while (scanf("%s%s", s1, s2))
    {
        PostOrder(s1, s2, strlen(s1));
        printf("\n");
    }
    return 0;
}

Python：

输入：

s = input()
s1, s2 = s.split(' ')

字符串截取：s[i,len]：从第i位（i>=0)开始到 len - 1 截取

代码：

def PostOrder(s1, s2, len1):
    if (len1 == 1):
        print(s1[0], end = '')
        return

    for i in range(len1):
        if s1[0] == s2[i]:
            if i != 0:
                PostOrder(s1[1: i + 1], s2[0: i], i)
            if i != len1 - 1:
                PostOrder(s1[i + 1: len1], s2[i + 1: len1], len1 - i - 1)
            break
    print(s1[0], end = '')

s = input()
s1, s2 = s.split(' ')
#print(s1[1:4], s2)
PostOrder(s1, s2, len(s1))