For example we have:

["p", "r", "e", "f", "e", "t", " ", "m", "a", "k", "e", " ", "p", "r", "a", "t", "i", " c", "e"]

We want to get:

["p", "r", "a", "t", "i", " c", "e", " ", "m", "a", "k", "e", " ", "p", "r", "e", "f", "e", "t"]

Requirement:

  You can only do in array swap, you cannot create a new array.

The way to do it:

  1. Reverse whole string to get:

["e", " c", "i", "t", "a", "r", "p", " ", "e", "k", "a", "m", " ", "t", "e", "f", "e", "r", "p"]

  2. Then we reverse each word to get final result:

["p", "r", "a", "t", "i", " c", "e", " ", "m", "a", "k", "e", " ", "p", "r", "e", "f", "e", "t"]

So the main function should looks like this:

let data = [
"p","r","e","f","e","t"," ",
"m","a","k","e"," ",
"p","r","a","t","i","c","e"
]; function main(_data) {
let data = _data.slice(); reverseWholeString(data);
reverseWords(data, ); return data;
} console.log(main(data));

The reverseWholeString function would be:

function reverseWholeString(data) {
let start = ,
end = data.length - ;
reverseChars(data, start, end);
} function reverseChars(data, start, end) {
while (start < end) {
[data[start], data[end]] = [data[end], data[start]];
start++;
end--;
}
}

reverseWords function would be:

function reverseWords(data, start) {
let index = findEmptyIndex(data, start);
let end = index - ; while (index !== -) {
reverseChars(data, start, end);
start = index + ;
index = findEmptyIndex(data, start);
end = index - ;
} reverseChars(data, start, data.length - );
} function findEmptyIndex(data, start) {
let index;
for (let i = start; i < data.length; i++) {
if (data[i] === " ") {
index = i;
break;
} else {
index = -;
}
} return index;
}

------------

Full code:

let data = [
"p",
"r",
"e",
"f",
"e",
"t",
" ",
"m",
"a",
"k",
"e",
" ",
"p",
"r",
"a",
"t",
"i",
" c",
"e"
]; function reverseWholeString(data) {
let start = 0,
end = data.length - 1;
reverseChars(data, start, end);
} function reverseChars(data, start, end) {
while (start < end) {
[data[start], data[end]] = [data[end], data[start]];
start++;
end--;
}
} function findEmptyIndex(data, start) {
let index;
for (let i = start; i < data.length; i++) {
if (data[i] === " ") {
index = i;
break;
} else {
index = -1;
}
} return index;
}
function reverseWords(data, start) {
let index = findEmptyIndex(data, start);
let end = index - 1; while (index !== -1) {
reverseChars(data, start, end);
start = index + 1;
index = findEmptyIndex(data, start);
end = index - 1;
} reverseChars(data, start, data.length - 1);
} function main(_data) {
let data = _data.slice(); reverseWholeString(data);
reverseWords(data, 0); return data;
} console.log(main(data));

  

[Algorithm] Reverse array of Chars by word的更多相关文章

  1. [Algorithm] Reverse a linked list

    It helps to understands how recursive calls works. function Node(val) { return { val, next: null }; ...

  2. [Algorithm] Chunk Array

    // --- Directions// Given an array and chunk size, divide the array into many subarrays// where each ...

  3. Reverse array

    数组颠倒算法 #include <iostream> #include <iterator> using namespace std; void reverse(int* A, ...

  4. reverse array java

    /* package whatever; // don't place package name! */ import java.util.*; import java.lang.*; import ...

  5. CUDA学习,使用shared memory实现Reverse Array

  6. java.lang.String

    1.String 是一个类,广泛应用于 Java 程序中,相当于一系列的字符串.在 Java 语言中 strings are objects.创建一个 strings 最直接的方式是 String g ...

  7. Java基础教程(20)--数字和字符串

    一.数字   在用到数字时,大多数情况下我们都会使用基本数据类型.例如: int i = 500; float gpa = 3.65f; byte mask = 0xff;   然而,有时候我们既需要 ...

  8. [Swift]LeetCode557. 反转字符串中的单词 III | Reverse Words in a String III

    Given a string, you need to reverse the order of characters in each word within a sentence while sti ...

  9. JavaScript Array 常用函数整理

    按字母顺序整理 索引 Array.prototype.concat() Array.prototype.filter() Array.prototype.indexOf() Array.prototy ...

随机推荐

  1. web项目启动执行方法

    近期在项目中需要将用户在web启动时就查询出来,当作缓存使用. 一.首先需要实现 ServletContextListener 接口 public class UserCacheUtils imple ...

  2. POJ 1222【异或高斯消元|二进制状态枚举】

    题目链接:[http://poj.org/problem?id=1222] 题意:Light Out,给出一个5 * 6的0,1矩阵,0表示灯熄灭,反之为灯亮.输出一种方案,使得所有的等都被熄灭. 题 ...

  3. logN判点是否在凸多边形内 HRBUSTOJ1429

    就是利用叉积的性质,如果向量A1到向量A2是顺时针则叉积为负反之为正. 然后我们可以二分的判断找到一个点恰被两条射线夹在一起. 然后我们再判断是否l,r这两个点所连直线与点的关系. 具体资料可以参照这 ...

  4. HDU 6134 Battlestation Operational(莫比乌斯反演)

    [题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=6134 [题目大意] 求$\sum_{i=1}^{n}{\sum_{j=1}^{i}\lceil{\ ...

  5. Android Studio NDK开发浅谈

    环境: Android Studio 1.1.0 NDK-r10d 1.新建项目--->包名:com.mxl.az.ndk 新建包含native方法的类:JniOperation.class p ...

  6. Mac 下解压NDK .bin文件

    Mac Android Studio 开发NDK,首先下载NDK文件----->android-ndk-r10d-darwin-x86_64.bin 1.打开终端获取文件权限 chmod a+x ...

  7. [CSAcademy]A-Game

    题目大意: 给你一个只含字符'A'和'B'的串,A和B两人轮流对其中的子串染色,要求被染色的子串中不包含已经被染色的子串. 最后,如果一方染的'A'少,那么这一方胜: 如果双方染的'A'和'B'一样多 ...

  8. 【8.31校内测试】【找规律二分】【DP】【背包+spfa】

    打表出奇迹!表打出来发现了神奇的规律: 1 1 2 2 3 4 4 4 5 6 6 7 8 8 8 8 9 10 10 11 12 12 12 13 14 14 15 16 16 16 16 16.. ...

  9. CROC 2016 - Elimination Round (Rated Unofficial Edition) E. Intellectual Inquiry 贪心 构造 dp

    E. Intellectual Inquiry 题目连接: http://www.codeforces.com/contest/655/problem/E Description After gett ...

  10. 2015 UESTC 数据结构专题G题 秋实大哥去打工 单调栈

    秋实大哥去打工 Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://acm.uestc.edu.cn/#/contest/show/59 Descr ...