【leetcode】第三章 哈希表part01

242.有效的字母异位词

• 使用HashMap

public boolean isAnagram(String s, String t) {HashMap<Character,Integer> map = new HashMap();int sLen = s.length();int tLen = t.length();if (sLen != tLen) return false;// 统计词频for (int i = 0; i < s.length(); i++) {char ch = s.charAt(i);map.put(ch,map.getOrDefault(ch,0)+1);}//输入: s = "anagram", t = "nagaram"//输出: true// 若某个字符频数小于0，则绝对不是异位词for (int i = 0; i < t.length(); i++) {char c = t.charAt(i);if (map.containsKey(c)) {if (map.get(c) <= 0) {return false;}map.put(c,map.get(c)-1);}else {return false;}}return true;
}

349. 两个数组的交集

• 使用map方法

public int[] intersection(int[] nums1, int[] nums2) {// 找到长度小的数组int size1 = nums1.length;int size2 = nums2.length;if (size1 > size2) {return intersectionNum(nums2,nums1);}else {return intersectionNum(nums1,nums2);}
}public int[] intersectionNum(int[] mapNum, int[] computeNum) {List<Integer> res = new ArrayList<>();HashMap<Integer,Integer> map = new HashMap<>();for (int i = 0; i < mapNum.length; i++) {map.put(mapNum[i],map.getOrDefault(mapNum[i],0)+1);}for (int i = 0; i < computeNum.length; i++) {if (map.containsKey(computeNum[i]) && map.get(computeNum[i])!=-1) {map.put(computeNum[i],-1);res.add(computeNum[i]);}}return res.stream().mapToInt(Integer::intValue).toArray();
}

• 使用set方法

public int[] intersection(int[] nums1, int[] nums2) {List<Integer> res = new ArrayList<>();HashSet<Integer> set = new HashSet<>();for(int num : nums1) {set.add(num);}for (int num :nums2) {if (set.contains(num)) {set.remove(num);res.add(num);}}return res.stream().mapToInt(Integer::intValue).toArray();}

202. 快乐数

public int getSum(int n) {int sum = 0;while (n != 0) {int bit = n % 10; // 9n = n / 10; // 1sum += Math.pow(bit,2); // 0 + 9}return sum;}
public boolean isHappy(int n) {// 无限循环代表会重复出现该sumHashSet<Integer> set = new HashSet<>();while (n != 1 && !set.contains(n)) {set.add(n);n = getSum(n);}return n == 1;
}

public int getSum(int n) {int sum = 0;while (n != 0) {int bit = n % 10; // 9n = n / 10; // 1sum += Math.pow(bit,2); // 0 + 9}return sum;}
public boolean isHappy(int n) {// 无限循环代表会重复出现该sumHashSet<Integer> set = new HashSet<>();while (true){if (n == 1) return true;if (set.contains(n)) return false;set.add(n);n = getSum(n);}
}

1. 两数之和

public int[] twoSum(int[] nums, int target) {HashMap<Integer,Integer> map = new HashMap<>();//输入：nums = [2,7,11,15], target = 9//输出：[0,1]//解释：因为 nums[0] + nums[1] == 9 ，返回 [0, 1] 。int[] res = new int[2];for (int i = 0; i < nums.length; i++) {if (map.containsKey(target-nums[i])) {res[0] = map.get(target-nums[i]);res[1] = i;}map.put(nums[i],i);}return res;}