3、哈希表

代码随想录
LeetCode 题解
查找表相关问题

1、有效的字母异位词

242 - 有效的字母异位词

public boolean isAnagram(String s, String t) {
    if (s.length() != t.length()) return false;
 
    int[] freq_s = new int[26];
    int[] freq_t = new int[26];
    for (char c : s.toCharArray()) freq_s[c - 'a']++;
    for (char c : t.toCharArray()) freq_t[c - 'a']++;
 
    for (int i = 0; i < 26; i++) {
        if (freq_s[i] != freq_t[i]) return false;
    }
    return true;
}

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

2、两个数组的交集

349 - 两个数组的交集

public int[] intersection(int[] nums1, int[] nums2) {
    Set<Integer> set = new HashSet<>();
    for (int i : nums1) set.add(i);
 
    Set<Integer> res = new HashSet<>();
    for (int i : nums2) {
        if (set.contains(i)) res.add(i);
    }
 
    int[] ret = new int[res.size()];
    int i = 0;
    for (int num : res) ret[i++] = num;
    return ret;
}

3、快乐数

202 - 快乐数

public boolean isHappy(int n) {
    // visit 中存放的是已经看过的数字
    Set<Integer> visit = new HashSet<>();
 
    while (n != 1 && !visit.contains(n)) {
        visit.add(n);
        n = getNextNumber(n);
    }
 
    return n == 1;
}
 
private int getNextNumber(int n) {
    int sum = 0;
    while (n != 0) {
        sum += (n % 10) * (n % 10);
        n /= 10;
    }
    return sum;
}

4、两数之和

1 - 两数之和
返回的是索引

public int[] twoSum(int[] nums, int target) {
    // num : index
    Map<Integer, Integer> map = new HashMap<>();
 
    for (int i = 0; i < nums.length; i++) {
        int need = target - nums[i];
        if (map.containsKey(need)) return new int[]{i, map.get(need)};
        map.put(nums[i], i);
    }
 
    throw new RuntimeException("无解");
}

5、四数相加 II

454 - 四数相加 II
返回和为 0 的个数

public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
    // sum : 个数
    Map<Integer, Integer> map = new HashMap<>();
    for (int i : nums1) {
        for (int j : nums2) {
            int sum = i + j;
            map.put(sum, map.getOrDefault(sum, 0) + 1);
        }
    }
 
    int res = 0;
    for (int k : nums3) {
        for (int l : nums4) {
            int sum = k + l;
            res += map.getOrDefault(-sum, 0);
        }
    }
 
    return res;
}

6、赎金信

383 - 赎金信

public boolean canConstruct(String ransomNote, String magazine) {
    if (ransomNote.length() > magazine.length()) return false;
 
    int[] freq_r = new int[26];
    int[] freq_m = new int[26];
    for (char c : ransomNote.toCharArray()) freq_r[c - 'a']++;
    for (char c : magazine.toCharArray()) freq_m[c - 'a']++;
 
    for (int i = 0; i < 26; i++) {
        if (freq_r[i] > freq_m[i]) return false;
    }
    return true;
}

7、三数之和

15 - 三数之和
返回和为 0 的三元组值 题解

public List<List<Integer>> threeSum(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>();
 
    // nums[i] + nums[p1] + nums[p2] = 0
    // nums[i] <= nums[p1] <= nums[p2]
    for (int i = 0; i < nums.length - 2; i++) {
        if (nums[i] > 0) break;
        if (i > 0 && nums[i] == nums[i - 1]) continue;
 
        int p1 = i + 1;
        int p2 = nums.length - 1;
        while (p1 < p2) {
            int v1 = nums[p1];
            int v2 = nums[p2];
            int sum = nums[i] + v1 + v2;
 
            if (sum < 0) {
                while (p1 < p2 && nums[p1] == v1) p1++;
            } else if (sum > 0) {
                while (p1 < p2 && nums[p2] == v2) p2--;
            } else {
                res.add(new ArrayList<Integer>(Arrays.asList(nums[i], v1, v2)));
                while (p1 < p2 && nums[p1] == v1) p1++;
                while (p1 < p2 && nums[p2] == v2) p2--;
            }
        }
    }
 
    return res;
}

8、四数之和

18 - 四数之和
返回和为 0 的四元组值

public List<List<Integer>> fourSum(int[] nums, int target) {
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>();
 
    // nums[i] + nums[j] + nums[p1] + nums[p2] = target
    // nums[i] <= nums[j] <= nums[p1] <= nums[p2]
    for (int i = 0; i < nums.length - 3; i++) {
        if (i > 0 && nums[i] == nums[i - 1]) continue;
 
        for (int j = i + 1; j < nums.length - 2; j++) {
            if (j > i + 1 && nums[j] == nums[j - 1]) continue;
 
            int p1 = j + 1;
            int p2 = nums.length - 1;
            while (p1 < p2) {
                long v1 = nums[p1];
                long v2 = nums[p2];
                long sum = nums[i] + nums[j] + v1 + v2;
 
                if (sum < target) {
                    while (p1 < p2 && nums[p1] == v1) p1++;
                } else if (sum > target) {
                    while (p1 < p2 && nums[p2] == v2) p2--;
                } else {
                    res.add(new ArrayList<>(Arrays.asList(nums[i], nums[j], (int) v1, (int) v2)));
                    while (p1 < p2 && nums[p1] == v1) p1++;
                    while (p1 < p2 && nums[p2] == v2) p2--;
                }
            }
        }
    }
 
    return res;
}

public List<List<Integer>> fourSum(int[] nums, int target) {
    int n = nums.length;
    Arrays.sort(nums);
    List<List<Integer>> res = new ArrayList<>();
 
    // nums[i] + nums[j] + nums[p1] + nums[p2] = target
    // nums[i] <= nums[j] <= nums[p1] <= nums[p2]
    for (int i = 0; i < nums.length - 3; i++) {
        if (i > 0 && nums[i] == nums[i - 1]) continue;
        // 提前终止
        if ((long) 4 * nums[i] > target) break;
        if ((long) nums[i] + nums[n - 1] + nums[n - 2] + nums[n - 3] < target) continue;
 
        for (int j = i + 1; j < nums.length - 2; j++) {
            if (j > i + 1 && nums[j] == nums[j - 1]) continue;
            // 提前终止
            if ((long) 3 * nums[j] + nums[i] > target) break;
            if ((long) nums[i] + nums[j] + nums[n - 1] + nums[n - 2] < target) continue;
 
            int p1 = j + 1;
            int p2 = nums.length - 1;
            while (p1 < p2) {
                long v1 = nums[p1];
                long v2 = nums[p2];
                long sum = nums[i] + nums[j] + v1 + v2;
 
                if (sum < target) {
                    while (p1 < p2 && nums[p1] == v1) p1++;
                } else if (sum > target) {
                    while (p1 < p2 && nums[p2] == v2) p2--;
                } else {
                    res.add(new ArrayList<>(Arrays.asList(nums[i], nums[j], (int) v1, (int) v2)));
                    while (p1 < p2 && nums[p1] == v1) p1++;
                    while (p1 < p2 && nums[p2] == v2) p2--;
                }
            }
        }
    }
 
    return res;
}