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Space Replacement

Source

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.

Example
Given "Mr John Smith", length = 13.

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

Note
If you are using Java or Python,please use characters array instead of string.

Challenge
Do it in-place.

题解

根据题意,给定的输入数组长度足够长,将空格替换为%20 后也不会溢出。通常的思维为从前向后遍历,遇到空格即将%20 插入到新数组中,这种方法在生成新数组时很直观,但要求原地替换时就不方便了,这时可联想到插入排序的做法——从后往前遍历,空格处标记下就好了。由于不知道新数组的长度,故首先需要遍历一次原数组,字符串类题中常用方法。

需要注意的是这个题并未说明多个空格如何处理,如果多个连续空格也当做一个空格时稍有不同。

Java

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) {
        if (string == null) return 0;

        int space = 0;
        for (char c : string) {
            if (c == ' ') space++;
        }

        int r = length + 2 * space - 1;
        for (int i = length - 1; i >= 0; i--) {
            if (string[i] != ' ') {
                string[r] = string[i];
                r--;
            } else {
                string[r--] = '0';
                string[r--] = '2';
                string[r--] = '%';
            }
        }

        return length + 2 * space;
    }
}

源码分析

先遍历一遍求得空格数,得到『新数组』的实际长度,从后往前遍历。

复杂度分析

遍历两次原数组,时间复杂度近似为 O(n)O(n), 使用了r 作为标记,空间复杂度 O(1)O(1).