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  • 6. ZigZag Conversion (字符串的连接)

    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 ''
        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]!=''){
            for(i = 0; i < numRows; i++){
                if(s[p]=='') break;
                g[i][j[i]++] = s[p++];
            }
            for(i = numRows-2; i >= 1; i--){
                if(s[p]=='') break;
                g[i][j[i]++] = s[p++];
            }
        }
    
        g[0][j[0]] = '';
        for(i = 1; i < numRows; i++){
            g[i][j[i]] = '';
            if(g[i][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;
    }
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  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/5362016.html
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