算法训练Day6|哈希表

242 有效的字母异位词

最简单的想法，排序后比较是否相等

public boolean isAnagram(String s, String t) {
    int sLength = s.length();
    int tLength = t.length();

    if (sLength != tLength) {
        return false;
    }

    s = s.chars()
            .sorted()
            .collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append)
            .toString();
    t = t.chars()
            .sorted()
            .collect(StringBuilder::new, StringBuilder::appendCodePoint, StringBuilder::append)
            .toString();

    if (s.equals(t)) {
        return true;
    }

    return false;

}

哈希表法

public boolean isAnagram(String s, String t) {
    int sLength = s.length();
    int tLength = t.length();

    if (sLength != tLength) {
        return false;
    }

    int[] hashMap = new int[26];

    for (int i = 0; i < sLength; i++) {
        hashMap[s.charAt(i) - 'a']++;
        hashMap[t.charAt(i) - 'a']--;
    }

    for (int i = 0; i < 26; i++) {
        if (hashMap[i]!= 0) {
            return false;
        }
    }

    return true;

}

349 两个数组的交集

public int[] intersection(int[] nums1, int[] nums2) {

    if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0) {
        return new int[0];
    }
    Set<Integer> set1 = new HashSet<>();
    Set<Integer> resultSet = new HashSet<>();
    //遍历数组1
    for (int i : nums1) {
        set1.add(i);
    }
    //遍历数组2的过程中判断哈希表中是否存在该元素
    for (int i : nums2) {
        if (set1.contains(i)) {
            resultSet.add(i);
        }
    }
    //将结果几何转为数组
    return resultSet.stream().mapToInt(x -> x).toArray();
}

202 快乐数

数字会循环，所以看set中是否存在曾经出现过的数字

private int getNext(int n) {
    int totalSum = 0;
    while (n > 0) {
        int d = n % 10;
        n = n / 10;
        totalSum += d * d;
    }

    return totalSum;
}
public boolean isHappy(int n) {
    Set<Integer> seen = new HashSet<>();

    while (n != 1 && !seen.contains(n)) {
        seen.add(n);
        n = getNext(n);
    }
    return n == 1;
}

双指针，判断是否有循环链表 快指针每次比慢指针多走一步，如果有环存在，那么肯定会有相等的时刻

private int getNext(int n) {
    int totalSum = 0;
    while (n > 0) {
        int d = n % 10;
        n = n / 10;
        totalSum += d * d;
    }

    return totalSum;
}
public boolean isHappy(int n) {
    int slow = n;
    int fast = getNext(n);

    while (fast != 1 && slow != fast) {
        slow = getNext(slow);
        fast = getNext(getNext(fast));
    }

    return fast == 1;
}

1 两数之和

a1 + a2 = target; 在hash表中寻找是否存在a1，使得target - a2 = a1

public int[] twoSum(int[] nums, int target) {
    
    Map<Integer,Integer> integerMap = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        if (integerMap.containsKey(target - nums[i])) {
            return new int[]{integerMap.get(target - nums[i]), i};
        }
        integerMap.put(nums[i], i);
    }
    return new int[0];
}