Leetcode c语言-ZigZag Conversion

Title:

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".


该题可能比较难理解,其实就是将一个字符串的序号重新排列成锯齿形。

比如一个字符串012345678910,row为3,那么应该如下排序:


从0到10的连线就是一个锯齿状。


那么这道题的解题思路就是找到每一行的间隔的规律,可以看出,每一行的最小间隔一定是2,然后每两个数字的间隔之和为row-1,这就是间隔的规律。知道这个规律后就很好办了:


Solution:

char* convert(char* s, int numRows) {
    int len = strlen(s);
    char* p = (char*)malloc(sizeof(char)*len+1);
    int i,j,m;
    int index =0;
    

    if ( numRows == 1 || len ==1 )
    {
        return s;
    }  
    
    for (j=0;j<len;j=j) {
        p[index++] = s[j];
        j=j+(numRows-1)*2;
        }
        
    for (m=1;m<numRows-1;m++) {
        for (i=j=m;j<len;j=j) {
        p[index++] = s[j];
            i = numRows-1-i;
            j=j+(i)*2;
        }
    }
    
    for (j=numRows-1;j<len;j=j) {
        p[index++] = s[j];
        j=j+(numRows-1)*2;
        }
    
    p[index] = 0;
    return p;
}


posted on 2017-09-10 10:20  sichenzhao  阅读(400)  评论(0编辑  收藏  举报

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