标签:string bsp abc ++ integer lower nal lis ict
question:
Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less 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 left the other as original. Example: Input: s = "abcdefg", k = 2 Output: "bacdfeg" Restrictions: The string consists of lower English letters only. Length of the given string and k will in the range [1, 10000]
try:
class Solution {
public String reverseStr(String s, int k) {
char[] charArray=s.toCharArray();
int length=charArray.length;
char[] result=new char[length];
for(int i=0;i<length;i++){
if((i/k)%2==0){
if(length<((i/k)*k+k)){
result[length-1-i%k]=charArray[i];
//continue;
}else{
result[(i/k)*k+k-1-i%k]=charArray[i];
}
}else {
result[i]=charArray[i];
}
}
return new String(result);
}
}
result: 22.58%
conclusion:
LeetCode 541. Reverse String II
标签:string bsp abc ++ integer lower nal lis ict
原文地址:https://www.cnblogs.com/hzg1981/p/8971683.html