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Leetcode 解题模版 - HashMap

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HashMap哈希表

HashMap

  1. 用Array+Linkedlist（chaining，另有openaddress的实现）实现的能在平均O(1)的时间内快速增删查的数据结构
  2. 表内存储的数据需要实现equals()和hashCode()

LinkedHashMap

  1. 有顺序的hashmap，遍历顺序是key插入的顺序
  2. 所有的key按顺序存成一个linkedlist

TreeMap

  1. 有顺序(sorted)的map，遍历顺序是key从小到大
  2. 所有的key存成一个红黑树（self-balanacing binary tree）
  3. 增删查O(logn)

HashSet

  1. 没有value的hashmap

适用场景(when to use)

当需要快速查找数据时，可以利用HashMap加速查找

注意是要查找的数据结构实现了equals()和hashCode()

Array 和 HashMap的区别在于：

   Array无法快速查找，HashMap可以

   Array里的元素是有顺序的，HashMap没有

   Array的overhead比较小，HashMap实现比较复杂

例题

Two Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

  • 2 <= nums.length <= 104
  • -109 <= nums[i] <= 109
  • -109 <= target <= 109
  • Only one valid answer exists.

Solution 1: Brute Force

class Solution {    public int[] twoSum(int[] nums, int target) {        for(int i = 0; i < nums.length; i++) {            for(int j = 0; j < nums.length; j++) {                if(nums[i] + nums[j] == target && i != j) {                    return new int[]{i, j};                }            }        }        return new int[]{-1,-1};    }}

Solution 2: 

当需要快速查找数据时，可以利用HashMap加速查找

查找对象：目前见过的值 - index

HashMap：

   记录下之前见过的{key: 值，value：index}

   对于array里的每一个数，如果map里存在target-x，那就找到了答案

Solution Step:

  1. Initialize HashMap (possible with initial size to improve performance)
  2. For each number x in array

       a. if target - x exists in array, return the indices

       b. put{x, index(x)}

时间复杂度：O(n)

空间复杂度：O(n)

class Solution {    public int[] twoSum(int[] nums, int target) {        Map<Integer, Integer> map = new HashMap(nums.length);  //when putting the initial size, the both runtime/memory increases significantly        for(int i = 0; i < nums.length; i++) {            if(map.containsKey(target - nums[i])) {                return new int[]{map.get(target - nums[i]),i};            } else {                map.put(nums[i],i);            }        }        return new int[]{-1,-1};    }}

Extention: 如果return的是值不是index的话：

Two Pointer（array needs to be sorted, or we sort by ourself），Opposite Direction

Each iteration we will have a sum of two pointers -> array[i] + array[j]

   case 1: array[i] + array[j] == target -> we found the answer

   case 2: array[i] + array[j] > target   -> too big, move the right pointer to left

   case 3: array[i] + array[j] < target   -> too small, move the left pointer to right

Solution:

  1. Sort the input array
  2. Initialize two pointers, i = 0 and j = n - 1
  3. while i < j:

        a. arr[i] + arr[j] == target --> return

        b. arr[i] + arr[j] >   target --> j--

        c. arr[i] + arr[j] <   target --> i++

class Solution {    public int[] twoSum(int[] nums, int target) {        Arrays.sort(nums);        int i = 0, j = nums.length - 1;        while(i < j) {            int sum = nums[i] + nums[j];            if(sum == target) {                return new int[]{nums[i], nums[j]};            } else if (sum > target) {                j--;            } else {                i++;            }        }        return null;}

560. Subarray Sum Equals K

Given an array of integers nums and an integer k, return the total number of subarrays whose sum equals tok.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [1,1,1], k = 2
Output: 2

Example 2:

Input: nums = [1,2,3], k = 3
Output: 2

Constraints:

  • 1 <= nums.length <= 2 * 104

  • -1000 <= nums[i] <= 1000

  • -107 <= k <= 107

Note: 所有的subarray sum都可以通过sum(0,x) - sum(0,y) 来计算

思路:

当需要快速查找数据时，可以利用HashMap加速查找

查找对象: all sum seen so far until current index -> # of times it occurs

当current sum - k 存在于map内，那么answer += map.get(sum - k)

因为sum - k存在于map说明之前sum(0, x) = k 出现了y次， 那么：

    subarray sum = current sum - sum(0,x) = k 多出现了y次

要点：Map内一开始存在一个{0, 1}, 代表sum = 0 默认出现了1次

Solution:

  1. Initialize HashMap with {0, 1}
  2. Initialize sum = 0, result = 0
  3. For each number x in array

        a. sum += x

        b. if sum - k in the map -> result += map.get(sum-k)

        c. put sum into the map, increase its count if exists

class Solution {    public int subarraySum(int[] nums, int k) {        Map<Integer, Integer> map = new HashMap(nums.length);        map.put(0,1);        int sum = 0, count = 0;        for(int x : nums) {            sum += x;            if(map.containsKey(sum -k)) {                count += map.get(sum-k);            }            map.put(sum, map.getOrDefault(sum,0)+1);        }        return count;    }}