一、242. 有效的字母异位词
题目连接：https://leetcode.cn/problems/valid-anagram/
思路：使用hash思想，将字母key转化为数组的索引，value存的每个字母的数量，在t中减去数量，在遍历数组，不等于0的即是不存在或者的少于字母
class Solution {
public boolean isAnagram(String s, String t) {
int record[] = new int[26];
for (int i = 0; i < s.length(); i++) {
record[s.charAt(i) - 'a']++;
}
for (int i = 0; i < t.length(); i++) {
record[t.charAt(i) - 'a']--;
}
for (int i = 0; i < record.length; i++) {
if (record[i] != 0) {
return false;
}
}
return true;
}
}
二、 383. 赎金信
题目连接：https://leetcode.cn/problems/ransom-note/
思路：先统计magszine里面的每个字符的的数量，在扫描ransomNote，在扫描的过程，数量--，如果每个字符的数量小于了0，则证明magazine里面的字符不够可以拼成ransomNote
class Solution {
public boolean canConstruct(String ransomNote, String magazine) {
int[] hash = new int[26];
for (int i = 0; i < magazine.length(); i++){
hash[magazine.charAt(i) - 'a']++;
}
for (int i = 0; i < ransomNote.length(); i++){
hash[ransomNote.charAt(i) - 'a']--;
if (hash[ransomNote.charAt(i) - 'a'] < 0) return false;
}
return true;
}
}
三、438. 找到字符串中所有字母异位词
题目连接：https://leetcode.cn/problems/find-all-anagrams-in-a-string/
1、暴力方法
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> result = new ArrayList<>();
int pLength = p.length();
int sLength = s.length();
for (int i = 0; i < sLength - pLength + 1; i++){
int hash[] = new int[26];
for (int k = 0; k < pLength; k++){
hash[p.charAt(k) - 'a']++;
}
for (int j = i; j < i + pLength ; j++) {
hash[s.charAt(j) - 'a']--;
}
boolean isHave = true;
for (int g = 0; g < hash.length; g++){
if (hash[g] != 0) {
isHave = false;
break;
}
}
if (isHave) result.add(i);
}
return result;
}
}
2、使用滑动窗口思想，不断修改窗口的起始位置统计数量，比较两个统计数量的集合是否相同
class Solution {
private boolean isArraysEquals(int[] sHash, int[] pHash) {
if (sHash == null || pHash == null) {
return false;
}
if (sHash.length != pHash.length) return false;
for (int i = 0; i < sHash.length; i++) {
if (sHash[i] != pHash[i]) return false;
}
return true;
}
public List<Integer> findAnagrams(String s, String p) {
List<Integer> result = new ArrayList<>();
int sLength = s.length();
int pLength = p.length();
if (sLength < pLength){
return result;
}
int[] sHash = new int[26];
int[] pHash = new int[26];
for (int i = 0; i < pLength; i++){
sHash[s.charAt(i) - 'a']++;
pHash[p.charAt(i) - 'a']++;
}
if (isArraysEquals(sHash, pHash)) {
result.add(0);
}
for (int i = pLength; i < sLength; i++){
sHash[s.charAt(i - pLength) - 'a']--;
sHash[s.charAt(i) - 'a']++;
if (isArraysEquals(sHash, pHash)){
result.add(i - pLength + 1);
}
}
return result;
}
}
四、349. 两个数组的交集
题目连接：https://leetcode.cn/problems/intersection-of-two-arrays/
思路：将nums1去重放入num1Set，遍历nums2,如果在num1Set,就放入结果集中
方法一、
class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
HashSet<Integer> num1Set = new HashSet<>();
for (int i = 0; i < nums1.length; i++){
num1Set.add(nums1[i]);
}
HashSet<Integer> resultHashSet = new HashSet<>();
for (int i = 0; i < nums2.length; i++){
if (num1Set.contains(nums2[i])) {
resultHashSet.add(nums2[i]);
}
}
int[] result = new int[resultHashSet.size()];
int index = 0;
Iterator<Integer> iterator = resultHashSet.iterator();
while (iterator.hasNext()) {
result[index++] = iterator.next();
}
return result;
}
}
方法二、
class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
int[] hash = new int[1001];
for (int i = 0; i < nums1.length; i++) {
hash[nums1[i]] = 1;
}
HashSet<Integer> resultHashSet = new HashSet<>();
for (int i = 0; i < nums2.length; i++){
if (hash[nums2[i]] == 1) resultHashSet.add(nums2[i]);
}
int[] result = new int[resultHashSet.size()];
Iterator<Integer> iterator = resultHashSet.iterator();
int index = 0;
while (iterator.hasNext()) {
result[index++] = iterator.next();
}
return result;
}
}
五、350. 两个数组的交集 II
题目连接：https://leetcode.cn/problems/intersection-of-two-arrays-ii/
方法一、
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
int[] hash = new int[1001];
for (int i = 0; i < nums1.length; i++){
hash[nums1[i]]++;
}
ArrayList<Integer> list = new ArrayList<>();
for (int i = 0; i < nums2.length; i++){
if (hash[nums2[i]] != 0) {
list.add(nums2[i]);
hash[nums2[i]]--;
}
}
int[] result = new int[list.size()];
for (int i = 0; i < result.length; i++){
result[i] = list.get(i);
}
return result;
}
}
方法二、
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
HashMap<Integer, Integer> hashMap = new HashMap<>();
for (int i = 0; i < nums1.length; i++){
hashMap.put(nums1[i], hashMap.getOrDefault(nums1[i], 0) + 1);
}
ArrayList<Integer> resultList = new ArrayList<>();
for (int i = 0; i < nums2.length; i++){
if (hashMap.getOrDefault(nums2[i], 0) != 0){
resultList.add(nums2[i]);
hashMap.put(nums2[i], hashMap.getOrDefault(nums2[i], 0) - 1);
}
}
int[] result = new int[resultList.size()];
for (int i = 0; i < resultList.size(); i++){
result[i] = resultList.get(i);
}
return result;
}
}
六、 49. 字母异位词分组
题目连接：https://leetcode.cn/problems/group-anagrams/
思路：使用hashmap key存排序后的str, value即异位词集合
class Solution {
public List<List<String>> groupAnagrams(String[] strs) {
HashMap<String, ArrayList<String>> hashmap = new HashMap<>();
for (int i = 0; i < strs.length; i++){
String el = strs[i];
char[] cs = el.toCharArray();
Arrays.sort(cs);
String elNew = String.valueOf(cs);
if (!hashmap.containsKey(elNew)) {
hashmap.put(elNew, new ArrayList<String>());
}
hashmap.get(elNew).add(el);
}
return new ArrayList(hashmap.values());
}
}
七、 202. 快乐数
题目连接：https://leetcode.cn/problems/happy-number/
思路：使用hashSet 判断和是否出现过，出现就终止判断是否n==1
class Solution {
public List<List<String>> groupAnagrams(String[] strs) {
HashMap<String, ArrayList<String>> hashMap = new HashMap<>();
for (String str:strs){
char[] cs = str.toCharArray();
Arrays.sort(cs);
String strSort = String.valueOf(cs);
if (hashMap.get(strSort) == null) {
hashMap.put(strSort, new ArrayList<String>());
}
hashMap.get(strSort).add(str);
}
return new ArrayList<>(hashMap.values());
}
}
八、1. 两数之和
题目连接：https://leetcode.cn/problems/two-sum/
思路一、暴力法（复杂度不过）
class Solution {
public int[] twoSum(int[] nums, int target) {
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
System.out.println(nums[i] + " " + nums[j]);
if (nums[i] + nums[j] == target) {
return new int[]{i, j};
}
}
}
return new int[]{-1, -1};
}
}
思路二、看target和当前nums[i]差值是否在hashMap中存在，存在则说明找到结果，如果不存在则将该nums[i]放入hashmap;
class Solution {
public int[] twoSum(int[] nums, int target) {
HashMap<Integer, Integer> hashMap = new HashMap<>();
for (int i = 0; i < nums.length; i++){
int temp = target - nums[i];
if (hashMap.containsKey(temp)) {
return new int[]{i, hashMap.get(temp)};
}
hashMap.put(nums[i], i);
}
return new int[]{-1, -1};
}
}
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