6. ZigZag Conversion （字符串的连接）

Posted 2020-07-06 joannae

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P A H N
A P L S I I G
Y I R
And then read line by line: "PAHNAPLSIIGYIR"
Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);
convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".

思路：Zigzag，一列长，一列短（少两个字符），长的那列由上到下写，短的由下到上写。

char* convert(char* s, int numRows) {
    int len = strlen(s);
    if(0 == len) return "";
   
    char g[numRows][len+1]; //+1 for ‘\0‘
    int p = 0; //pointer to s
    int i = 0; //line number
    int j[numRows]; //column number
    for(i = 0; i < numRows; i++){
        j[i] = 0;
    }

    while(s[p]!=‘\0‘){
        for(i = 0; i < numRows; i++){
            if(s[p]==‘\0‘) break;
            g[i][j[i]++] = s[p++];
        }
        for(i = numRows-2; i >= 1; i--){
            if(s[p]==‘\0‘) break;
            g[i][j[i]++] = s[p++];
        }
    }

    g[0][j[0]] = ‘\0‘;
    for(i = 1; i < numRows; i++){
        g[i][j[i]] = ‘\0‘;
        if(g[i][0]!=‘\0‘) strcat(g[0], g[i]);
    }
        
    //g[0] is saved in stack, which will be deleted when leave the function so we should copy it before leave
    char* ret = malloc(sizeof(char)*(len+1)); //saved in heap, which won‘t be disappeared when leave the function
    memcpy(ret, g[0], sizeof(char)*(len+1));
    return ret;
}