标签：public count eof ring rest str swap 字符 value
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]
//每2k个字符里，翻转前k个
class Solution {
public String reverseStr(String s, int k) {
char[] arr = s.toCharArray();
int n = arr.length;
int i = 0;
while (i < n) {
int j = Math.min(i + k - 1, n - 1); //i的值确定后，j的值为i + k - 1，要判断是否超出数组(和n - 1比较)
swap(arr, i, j);
i += 2 * k;
}
return new String(arr);
//return String.valueOf(arr);
}
private void swap(char[] arr, int l, int r) {
while (l < r) {
char temp = arr[l];
arr[l++] = arr[r];
arr[r--] = temp;
}
}
}
541. Reverse String II
标签：public count eof ring rest str swap 字符 value
原文地址：https://www.cnblogs.com/MarkLeeBYR/p/10678577.html
