代码随想录四刷day4

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前言

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哈希表中关键码就是数组的索引下标，然后通过下标直接访问数组中的元

一、力扣面试题 02.07. 链表相交

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        ListNode L1, L2, HA = new ListNode(-1,headA), HB = new ListNode(-1,headB);
        ListNode p1 = HA, p2 = HB;
        int lenA = 0, lenB = 0, margin = 0;
        while(p1.next != null){
            lenA ++;
            p1 = p1.next;
        }
        while(p2.next != null){
            lenB ++;
            p2 = p2.next;
        }
        if(lenA > lenB){
            L1 = HA;
            L2 = HB;
            margin = lenA - lenB;
        }else{
            L1 = HB;
            L2 = HA;
            margin = lenB - lenA;
        }
        while(margin > 0){
            margin --;
            L1 = L1.next;
        }
        while(L1 != null && L2 != null){
            if(L1 == L2){
                return L1;
            }
            L1 = L1.next;
            L2 = L2.next;
        }
        return null;
    }
}

二、力扣142. 环形链表 II

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if(head == null || head.next == null){
            return null;
        }
        ListNode L = new ListNode(-1,head);
        ListNode fast = L.next.next, slow = L.next;
        while(fast != null){
            if(fast == slow){
                ListNode p = L;
                while(slow != p){
                    slow = slow.next;
                    p = p.next;
                }
                return p;
            }else{
                if(fast.next == null || fast.next.next == null){
                    return null;
                }
                fast = fast.next.next;
                slow = slow.next;
            }
        }
        return null;
    }
}

三、力扣242. 有效的字母异位词

class Solution {
    public boolean isAnagram(String s, String t) {
        int[] flagA = new int[27];
        int[] flagB = new int[27];
        for(char c : s.toCharArray()){
            flagA[c-'a'] += 1;
        }
        for(char c : t.toCharArray()){
            flagB[c - 'a'] += 1;
        }
        for(int i = 0; i < 27; i ++){
            if(flagA[i] != flagB[i]){
                return false;
            }
        }
        return true;
    }
}

四、力扣9. 字母异位词分组

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        List<List<String>> res = new ArrayList<>();
        Map<String,List<String>> map = new HashMap<>();
        for(String s : strs){
            String t = fun(s);
            if(!map.containsKey(t)){
                map.put(t,new ArrayList<>());
            }
            List<String> li = map.get(t);
            li.add(s);
            map.put(t,li);
        }
        res.addAll(map.values());
        return res;
    }
    public String fun(String s){
        String res = "";
        int[] arr = new int[27];
        for(char c : s.toCharArray()){
            arr[c-'a'] += 1;
        }
        for(int i = 0; i< 26; i ++){
            if(arr[i] != 0){
                char c =(char)  ('a' + i);
                res += c;
                res += arr[i];
            }
        }
        return res;
    }
}

五、力扣9. 字母异位词分组

class Solution {
    public List<List<String>> groupAnagrams(String[] strs) {
        List<List<String>> res = new ArrayList<>();
        Map<String,List<String>> map = new HashMap<>();
        for(String s : strs){
            StringBuilder sb = new StringBuilder();
            int[] arr = new int[26];
            for(char c : s.toCharArray()){
                arr[c-'a'] ++;
            }
            for(int i = 0; i < 26; i ++){
                if(arr[i] != 0){
                    sb.append('a'+i).append(arr[i]);
                }
            }
            String t = sb.toString();
            map.put(t,map.getOrDefault(t,new ArrayList<>()));
            map.get(t).add(s);
        }
        res.addAll(map.values());
        return res;
    }
}