题目:ZigZag Conversion
一串字符按照Z字形的数组给了我们,要求转成原本的顺序。
思路:
统计“|/”的个数;
竖着的和斜着的下标有对应关系;
竖着的:k = j*(2*numRows - 2) + i;
斜着的(不含两个端点):k = (j + 1)*(2*numRows - 2) - i;
注意:可能会有残缺的部分。
/****************************************************************** ZigZag Conversion 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". ******************************************************************/ #include <stdio.h> #include <memory.h> char* convert(char* s, int numRows) { int length = strlen(s); printf("%d ",length); if(length <= numRows || numRows <= 1)return s; int zn = length/(2*numRows - 2);//Z字形竖着放时,最开始的一竖和一撇的组合整体的个数 int sn = length%(2*numRows - 2);//Z字形竖着放时,剩下残缺的组合的字母个数 char *cs = (char *)malloc((length + 1)*sizeof(char)); memset(cs,0,(length + 1)*sizeof(char)); int index = 0,k = 0; for (int i = 0;i < numRows;i++){//完整组合的对应转换 for(int j = 0;j < zn;j++){//竖线上的点坐标对应公式 k = j*(2*numRows - 2) + i; cs[index++] = s[k]; if(i > 0 && i < numRows - 1){//斜线上的不含两端点的点坐标对应公式 k = (j + 1)*(2*numRows - 2) - i; cs[index++] = s[k]; } } if(sn > i){//残缺组合的对应转换 k = zn*(2*numRows - 2) + i; cs[index++] = s[k]; if(i > 0 && i < numRows - 1 && sn > 2*numRows - 2 - i){ k = (zn + 1)*(2*numRows - 2) - i; cs[index++] = s[k]; } } } return cs; } void main(){ char s[] = "Apalindromeisaword,phrase,number,orothersequenceofunitsthatcanbereadthesamewayineitherdirection,withgeneralallowancesforadjustmentstopunctuationandworddividers."; char *cs = convert(s,2); printf("%s ",cs); free(cs); }