資訊專(zhuān)欄INFORMATION COLUMN
Problem
Given a string and an offset, rotate string by offset. (rotate from left to right)
ExampleGiven "abcdefg".
offset=0 => "abcdefg"
offset=1 => "gabcdef"
offset=2 => "fgabcde"
offset=3 => "efgabcd"
Rotate in-place with O(1) extra memory.
Note完成challenge，三步翻轉(zhuǎn)法。見(jiàn)Sol.2。
Solution1.
public class Solution {
public void rotateString(char[] str, int offset) {
int len = str.length;
char[] newstr = new char[len+1];
if (str == null || str.length == 0) return;
//modify offset if offset > length;
offset = offset % len;
for (int i = 0; i < offset; i++) {
newstr[i] = str[len-offset+i];
}
for (int i = offset; i < len; i++) {
newstr[i] = str[i-offset];
}
for (int i = 0; i < len; i++) {
str[i] = newstr[i];
}
return;
}
}
三步翻轉(zhuǎn)法：O(1) extra memory
public class Solution {
public void rotateString(char[] str, int offset) {
if (str == null || str.length == 0) {
return;
}
int len = str.length;
offset = offset % len;
reverse(str, 0, len-offset-1);
reverse(str, len-offset, len-1);
reverse(str, 0, len-1);
return;
}
public void reverse(char[] str, int start, int end) {
while (start <= end) {
char temp = str[start];
str[start] = str[end];
str[end] = temp;
start++;
end--;
}
}
}
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