剑指offer String

剑指offer-替换空格

Posted by franki on April 20, 2021

替换空格

实现一个函数，把字符串中的每个空格替换成“%20”。例如”we are happy.”，则输出”we%20are%20happy.”

思路

两种方案：

  1. 在原来的字符串上进行替换，可能覆盖修改在该字符串后面的内存。
  2. 创建新的字符串并在新的字符串上进行替换，可以分配足够多的内存。(计算出当前字符串存在的空格数量，需要开辟新的字符空间为originalLength + spaceLength*2,往后向前遍历，遇到空格在相应的位置加入”%20”，即是所求)

代码

function replaceSpace(string) {
    if (string.length === 0 || !string) {
        return;
    }
    let originalLength = string.length;
    let spaceLength = 0;
    let index = 0;
    while (index < originalLength) {
        if (string[index] === " ") {
            spaceLength++;
        }
        index++;
    }
    let newLength = originalLength + spaceLength * 2;
    let indexOfOriginal = originalLength;
    let indexOfNew = newLength;
    let newString = new Array(newLength);
    while (indexOfOriginal >= 0 && indexOfNew >= indexOfOriginal) {
        if (string[indexOfOriginal - 1] === " ") {
            newString[--indexOfNew] = '0';
            newString[--indexOfNew] = '2';
            newString[--indexOfNew] = '%';
        } else {
            newString[--indexOfNew] = string[indexOfOriginal - 1];
        }
        indexOfOriginal--;
    }

    return newString.join("");
}

复杂度分析

  • 时间复杂度: O(N)
  • 空间复杂度: O(N)
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