力扣---2020.3.16

面试题 01.06. 字符串压缩

class Solution {
    public String compressString(String S) {
        int N = S.length();
        int i = 0;
        StringBuilder sb = new StringBuilder();
        while (i < N) {
            int j = i;
            while (j < N && S.charAt(j) == S.charAt(i)) {
                j++;
            }
            sb.append(S.charAt(i));
            sb.append(j - i);
            i = j;
        }

        String res = sb.toString();
        return res.length() < S.length()? res:S;
    }
}

面试题58 - II. 左旋转字符串

//substring(int beginIndex, int endIndex)  方法返回字符串的子字符串。
class Solution {
    public String reverseLeftWords(String s, int n) {
        return s.substring(n) + s.substring(0, n);
    }
}

class Solution {
    public String reverseLeftWords(String s, int n) {
        char[] chars = s.toCharArray();
        reverse(chars, 0, n-1);
        reverse(chars, n, chars.length-1);
        reverse(chars, 0, chars.length-1);
        return new String(chars);
    }
    private void reverse(char[] arr, int i, int j) {
        while(i < j) {
            char x = arr[i];
            arr[i] = arr[j];
            arr[j] = x;
            i++;
            j--;
        }
    }
}

class Solution {
    public String reverseLeftWords(String s, int n) {
        String sub = s.substring(n);
        StringBuilder sb = new StringBuilder();
        sb.append(sub);

        String sub2 = s.substring(0, n);
        sb.append(sub2);

        return sb.toString();
    }
}

class Solution {
    public String reverseLeftWords(String s, int n) {
        char [] res = new char[s.length()];
        char [] temp = s.toCharArray();
        System.arraycopy(temp,0,res,s.length() - n,n);
        System.arraycopy(temp,n,res,0,s.length() - n);
        return String.copyValueOf(res);
    }
}

• system提供了一个静态方法arraycopy(),我们可以使用它来实现数组之间的复制.

面试题22. 链表中倒数第k个节点

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
        if(head == null||head.next == null){
            return head;
        }
        ListNode curr = head;
        while( k>0 && curr != null){
            curr = curr.next;
            k--;
        }
        while(curr != null){
            curr = curr.next;
            head = head.next;
        }
        return head;
    }
}

class Solution {
    public ListNode getKthFromEnd(ListNode head, int k) {
       ListNode fast = head;
        while(fast!=null) {
        	fast = fast.next;
        	if(k==0) {
        		head = head.next;
        	}else {
        		k--;
        	}
        }
        return head; 
    }
}

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