lintcode-easy-Space Replacement-优快云博客

本文介绍了一种将字符串中的所有空格替换为%20的方法。该方法适用于字符数组形式的字符串，并假设数组有足够的空间进行替换操作。文章提供了一个Java实现示例，详细解释了如何计算空格数量并进行替换，最后返回替换后的字符串长度。

Write a method to replace all spaces in a string with%20. The string is given in a characters array, you can assume it has enough space for replacement and you are given the true length of the string.

You code should also return the new length of the string after replacement.

Given "Mr John Smith", length = 13.

The string after replacement should be "Mr%20John%20Smith".

public class Solution {
    /**
     * @param string: An array of Char
     * @param length: The true length of the string
     * @return: The true length of new string
     */
    public int replaceBlank(char[] string, int length) {
        // Write your code here
        if(string == null || string.length == 0)
            return 0;
        
        int space_count = 0;
        int j = length - 1;
        
        while(j >= 0){
            if(string[j] == ' ')
                space_count++;
            j--;
        }
        
        int i = length + space_count * 2 - 1;
        
        for(j = length - 1; j >= 0; j--){
            if(string[j] == ' '){
                string[i--] = '0';
                string[i--] = '2';
                string[i--] = '%';
            }
            else{
                string[i--] = string[j];
            }
        }
        
        return length + space_count * 2;
    }
}