344. Reverse String
Write a function that reverses a string. The input string is given as an array of characters s.
You must do this by modifying the input array in-place with O(1) extra memory.
Example 1:
Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]
Example 2:
Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]
Constraints:
1 <= s.length <= 105s[i]is a printable ascii character.
没什么说的，左右双指针对向移动，交换元素即可。
代码：
class Solution {
public void reverseString(char[] s) {
if(s.length == 1) {
return;
}
int left = 0;
int right = s.length - 1;
while(left < right) {
char c = s[left];
s[left] = s[right];
s[right] = c;
left++;
right--;
}
return;
}
}
541. Reverse String II
Given a string s and an integer k, reverse the first k characters for every 2k characters counting from the start of the string.
If there are fewer than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and leave the other as original.
Example 1:
Input: s = "abcdefg", k = 2
Output: "bacdfeg"
Example 2:
Input: s = "abcd", k = 2
Output: "bacd"
Constraints:
1 <= s.length <= 104sconsists of only lowercase English letters.1 <= k <= 104
反转字符串的变形，控制好边界条件即可。 代码：
class Solution {
public String reverseStr(String s, int k) {
int len = s.length();
char[] ch = s.toCharArray();
int left = 0;
int right = len < k ? len - 1 : k - 1;
int offset = 0;
while(offset < len) {
left = offset;
right = offset + k > len ? len - 1: offset + k - 1;
while(left < right) {
char tc = ch[left];
ch[left] = ch[right];
ch[right] = tc;
left++;
right--;
}
offset += 2*k;
}
String ans = new String(ch);
return ans;
}
}
剑指 Offer 05. 替换空格
请实现一个函数，把字符串 s 中的每个空格替换成"%20"。
示例 1：
输入： s = "We are happy."
输出： "We%20are%20happy."
限制：
0 <= s 的长度 <= 10000
剑指OFFER系列英文版力扣没有。题目没什么说的，StringBuilder append很轻松。回头要看一下不用StringBuilder的双指针法。 代码：
class Solution {
public String replaceSpace(String s) {
StringBuilder sb = new StringBuilder();
for(char c : s.toCharArray()) {
if(c == ' ') {
sb.append("%20");
}
else {
sb.append(c);
}
}
return sb.toString();
}
}
151. Reverse Words in a String
Given an input string s, reverse the order of the words.
A word is defined as a sequence of non-space characters. The words in s will be separated by at least one space.
Return a string of the words in reverse order concatenated by a single space.
Note that s may contain leading or trailing spaces or multiple spaces between two words. The returned string should only have a single space separating the words. Do not include any extra spaces.
Example 1:
Input: s = "the sky is blue"
Output: "blue is sky the"
Example 2:
Input: s = " hello world "
Output: "world hello"
Explanation: Your reversed string should not contain leading or trailing spaces.
Example 3:
Input: s = "a good example"
Output: "example good a"
Explanation: You need to reduce multiple spaces between two words to a single space in the reversed string.
Constraints:
1 <= s.length <= 104scontains English letters (upper-case and lower-case), digits, and spaces' '.- There is at least one word in
s.
这题费了好大劲儿，二刷要注意。主要卡在Remove Extra Space功能上没想清楚。快慢指针还是不熟练。 代码：
class Solution {
public char[] removeExtraSpaces(char[] ch) {
int slow = 0;
for(int i=0; i<ch.length; i++) {
if(ch[i] != ' ') {
if(slow != 0) {
ch[slow++] = ' ';
}
while(i<ch.length && ch[i] != ' ') {
ch[slow++] = ch[i++];
}
}
}
ch = Arrays.copyOf(ch, slow);
return ch;
}
public String reverseWords(String s) {
char[] ch = s.toCharArray();
int left = 0;
int right = ch.length - 1;
//while(ch1[left] == ' ') left++;
//while(ch1[right] == ' ') right--;
//System.out.println("left:" + left + ", right: " + right);
//left = 0;
//right = 0;
ch = removeExtraSpaces(ch);
//System.out.println(ch);
left = 0;
right = ch.length - 1;
while(left < right) {
char tc = ch[left];
ch[left] = ch[right];
ch[right] = tc;
left++;
right--;
}
left=0;
right=0;
for(int i=0; i<=ch.length; i++) {
if(i == ch.length || ch[i] == ' ') {
right = i - 1;
while(left < right) {
char tc = ch[left];
ch[left] = ch[right];
ch[right] = tc;
left++;
right--;
}
left = i + 1;
}
}
String ans = new String(ch);
return ans;
}
}
剑指 Offer 58 - II. 左旋转字符串
示例 1：
输入: s = "abcdefg", k = 2
输出: "cdefgab"
示例 2：
输入: s = "lrloseumgh", k = 6
输出: "umghlrlose"
限制：
1 <= k < s.length <= 10000
对Java来说这题总归做不成O(1)的空间复杂度。 看了Carl的写法才来写的。三次旋转很巧妙。 前部分反转，后部分反转，再整体反转，也能达到左旋的效果。
class Solution {
public String reverseLeftWords(String s, int n) {
String str1 = s.substring(0,n);
String str2 = s.substring(n, s.length());
StringBuilder sb1 = new StringBuilder(str1).reverse();
str2 = new StringBuilder(str2).reverse().toString();
sb1.append(str2);
sb1.reverse();
str1 = sb1.toString();
return str1;
}
}
