替换空格

题目描述

请实现一个函数,将一个字符串中的每个空格替换成“%20”。例如,当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy。

思路: 预先计算出替换后的字符串的长度, 然后从后向前复制

版本一: 没做出来, 参考书上解题思路

class Solution {
public:
	void replaceSpace(char *str,int length) {
                int numBef = 0;
        int numAft = 0;
        int numBank = 0;
        int i = 0;

        while ('\0' != str[i]) {
            numBef++;
            if (' ' == str[i]) {
                numBank++;
            }
            i++;
        }

        numBef += 1;
        numAft = numBef + numBank * 2;

        while (0 <= numBef) {
            if (' ' == str[numBef]) {
                str[numAft] = '0';
                str[numAft - 1] = '2';
                str[numAft - 2] = '%';
                numBef--;
                numAft -= 3;
                continue;
            }
            str[numAft] = str[numBef];
            numBef--;
            numAft--;
        }
	}
};

版二: 书上改了不点不点

class Solution {
public:
	void replaceSpace(char *str,int length) {
        if ((NULL == str) || (0 == length)) {
            return;
        }
        
        int originalLength = 0;
        int numberOfBank = 0;
        int i = 0;
        while ('\0' != str[i]) {
            if (' ' == str[i]) {
                numberOfBank++;
            }
            originalLength++;
            i++;
        }
        
        int newLength = originalLength + numberOfBank * 2;
        
        if (newLength > length) {
            return;
        }
        
        int oldIndex = originalLength;
        int newIndex = newLength;
        
        while((oldIndex >= 0) && (newIndex > oldIndex)) {
            if (' ' == str[oldIndex]) {
                str[newIndex--] = '0';
                str[newIndex--] = '2';
                str[newIndex--] = '%';
            }
            else {
                str[newIndex--] = str[oldIndex];
            }
            oldIndex--;
        }
	}
};
posted @ 2019-02-20 22:56  张飘扬  阅读(120)  评论(0编辑  收藏  举报