代码随想录算法训练营第6天

class Solution {
    public boolean isAnagram(String s, String t) {
        int[] hash_word = new int[26];
        for (int i = 0; i < s.length(); i++) {
            hash_word[s.charAt(i) - 'a'] += 1;
        }
        for (int i = 0; i < t.length(); i++) {
            hash_word[t.charAt(i) - 'a'] -= 1;
        }
        for (int i = 0; i < hash_word.length; i++) {
            if (hash_word[i] != 0) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set1 = new HashSet<Integer>();
        Set<Integer> set2 = new HashSet<Integer>();
        for (int i = 0; i < nums1.length; i++) {
            set1.add(nums1[i]);
        }
        for (int i = 0; i < nums2.length; i++) {
            if (set1.contains(nums2[i])) {
                set2.add(nums2[i]);
            }
        }
 
        int[] ans = new int[set2.size()];
        int i = 0;
        for (int elem : set2) {
            ans[i] = elem;
            i++;
        }
        return ans;
    }
}

class Solution {
    public boolean isHappy(int n) {
        Set<Integer> tmp_resu = new HashSet<Integer>();
        int now_sum = squareSum(n);
        while (now_sum != 1) {
            if (tmp_resu.contains(now_sum) == false) {
                tmp_resu.add(now_sum);
            }
            else {
                return false;
            }
            now_sum = squareSum(now_sum);
        }
        return true;
    }
 
    public int squareSum(int n) {
        int sum = 0;
        while (n > 0) {  // 计算平方和      
            sum += (n % 10) * (n % 10);
            n = n / 10;
        }
        System.out.println(sum);
        return sum;
    }
}

class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> scaned_nums = new HashMap<Integer, Integer>();
        for (int i = 0; i < nums.length; i++) {
            int need = target - nums[i];
            if (scaned_nums.containsKey(need)) {
                return new int[]{i, scaned_nums.get(need)};
            }
            else {
                scaned_nums.put(nums[i], i);
            }
        }
        // 根据题目描述，一定有解；
        // 但对于程序来讲，必须保证每个分支都有return
        // 故写一个[-1, -1]，但这句是永远不可能走到的
        return new int[]{-1, -1};
    }
}