剑指Offer之Java算法习题精讲链表与字符串及数组

2024-09-30 17:01:33

题目一

解法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeElements(ListNode head, int val) {
        ListNode q = new ListNode(-1);
        ListNode a = q;
        q.next = head;
        while(q.next!=null){
            if(q.next.val==val){
                q.next = q.next.next;
            }else{
                q = q.next;
            }
        }
        return a.next;
    }
}

题目二

解法

class Solution {
    public boolean isIsomorphic(String s, String t) {
        if(s.length()!=t.length()){
            return false;
        }
        for(int i =0;i<s.length();i++){
            if(s.indexOf(s.charAt(i))!=t.indexOf(t.charAt(i))){
                return false;
            }
        }
        return true;
    }
}

题目三

解法

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void deleteNode(ListNode node) {
        node.val = node.next.val;
        node.next = node.next.next;
    }
}

题目四

解法

class Solution {
    public boolean isAnagram(String s, String t) {
        if(s.length()!=t.length()) return false;
        ArrayList<Character> list1 = new ArrayList<>();
        ArrayList<Character> list2 = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            list1.add(s.charAt(i));
        }

        for (int i = 0; i < t.length(); i++) {
            list2.add(t.charAt(i));
        }
        Collections.sort(list1);
        Collections.sort(list2);
        for (int i = 0; i < list1.size(); i++) {
            if (list1.get(i)!=list2.get(i)) return false;
        }
        return true;
    }
}

//要熟悉API
class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) {
            return false;
        }
        char[] str1 = s.toCharArray();
        char[] str2 = t.toCharArray();
        Arrays.sort(str1);
        Arrays.sort(str2);
        return Arrays.equals(str1, str2);
    }
}

题目五

解法

class Solution {
    public int missingNumber(int[] nums) {
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++) {
            if(i!=nums[i]){
                return i;
            }
        }
        return nums.length;
    }
}

题目六

解法

class Solution {
    public boolean wordPattern(String pattern, String s) {
        String[] split = s.split(" ");
        if(split.length!=pattern.length()) return false;
        int q = 0;
        for(int i =0;i<split.length;i++){
           l: for(int w = 0;w<split.length;w++){
                if(split[w].equals(split[i])){
                   q = w;
                   break l;
                }
            }
            if(pattern.indexOf(pattern.charAt(i))!=q){
                return false;
            }
        }
        return true;
    }
}

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剑指Offer之Java算法习题精讲链表与字符串及数组

题目一

题目二

题目三

题目四

题目五

题目六

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