JavaⅡ.左旋转字符串(substirng、列表遍历拼接、字符串遍历拼接)

剑指offer58. Ⅱ.左旋转字符串

解题思路1:
使用s.substring()方法

class Solution {
    public String reverseLeftWords(String s, int n) {
        return s.substring(n) + s.substring(0, n);
    }
}

解题思路2:
列表遍历拼接

class Solution {
    public String reverseLeftWords(String s, int n) {
        StringBuilder sb = new StringBuilder();
        for(int i = n; i < s.length(); i++){
            sb.append(s.charAt(i));
        }
        for(int i = 0; i < n; i++){
            sb.append(s.charAt(i));
        }
        return sb.toString();
    }
}

注意:
如图上红框所示, 题中已给范围是 k < s.length, 故不用考虑n大于字符串长度的情况
可将上面代码用%简化

//该代码同样适用于 n > s.length的情况
class Solution {
    public String reverseLeftWords(String s, int n) {
        StringBuilder sb = new StringBuilder();
        for(int i = n; i < n + s.length(); i++){
            sb.append(s.charAt(i % s.length()));
        }
        return sb.toString();
    }
}

解题思路3:
字符串遍历拼接

class Solution {
    public String reverseLeftWords(String s, int n) {
        String str = "";
        for(int i = n; i < s.length(); i++){
            str += s.charAt(i);
        }
        for(int i = 0; i < n; i++){
            str += s.charAt(i);
        }
        return str;
    }
}

用%简化

class Solution {
    public String reverseLeftWords(String s, int n) {
        String str = "";
        for(int i = n; i < n + s.length(); i++){
            str += s.charAt(i % s.length());
        }
        return str;
    }
}