剑指Offer之Java算法习题精讲字符串与二叉搜索树

题目一

解法

class Solution {
    public boolean repeatedSubstringPattern(String a) {
        for (int i = 1; i <=a.length()/2 ; i++) {
            String s = a.substring(0, i);
            StringBuffer sb = new StringBuffer();
            while (sb.length() 
 

题目二

解法

class Solution {
    public boolean detectCapitalUse(String word) {
        if(word.toLowerCase().equals(word)||word.toUpperCase().equals(word)||word.substring(1, word.length()).toLowerCase().equals(word.substring(1, word.length()))) return true;
        return false;
    }
}

题目三

解法

二叉搜索树有个性质为二叉搜索树中序遍历得到的值序列是递增有序的

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int ans;
    int pre;
    public int getMinimumDifference(TreeNode root) {
        ans = Integer.MAX_VALUE;
        pre = -1;
        method(root);
        return ans;
    }
    public void method(TreeNode root){
        if(root==null){
            return;
        }
        method(root.left);
        if(pre==-1){
            pre = root.val;
        }else{
            ans = Math.min(ans, root.val - pre);
            pre = root.val;
        }
        method(root.right);
    }
}

题目四

 解法

class Solution {
    public String reverseStr(String a, int k) {
        int con = 1;
        StringBuffer sb = new StringBuffer();
        while (con*k<=a.length()){
            String substring = a.substring((con - 1) * k, con * k);
            if(con%2==0){
                sb.append(substring);
                con++;
            }else {
                for (int i1 = substring.length()-1; i1 >=0 ; i1--) {
                    sb.append(substring.charAt(i1));
                }
                con++;
            }
        }
        if((con-1)*k=0; i1--) {
                    sb.append(s.charAt(i1));
                }
            }else {
                sb.append(s);
            }
        }
        return sb.toString();
    }
}

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