HashMap 专题

hashmap map key and value for highly efficient lookup.

hashmap search used binary tree O(log2)

main methods

int size();

hashmap.put(key,value);

Value get(key);

boolean containsKey(key);

boolean containsValue(Value);

remove（key）

leetcode：

1.Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

solution:

Key:value

Value:indice

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer,Integer> hashmap=new  HashMap<Integer,Integer>();
        int[] res=new int[2];
        int l=nums.length;
        for(int i=0;i<l;i++){
            if(hashmap.containsKey(target-nums[i])){
                res[0]=hashmap.get(target-nums[i])+1;
                res[1]=i+1;
            }
            else{
                hashmap.put(nums[i],i);
            }
        }
        return res;
    }
}

 

Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

public class Solution {
    public int lengthOfLongestSubstring(String s) {
    if(s.length()<=1){
            return s.length();
        }
    HashMap<Character,Integer> map=new HashMap<Character,Integer>();
    int max=0;
    int current=0;
    char[] c=s.toCharArray();
    for(int i=0;i<c.length;i++){
        char v=c[i];
        if(map.containsKey(v)&&map.get(v)>=current){
            current=map.get(v)+1;
        }
        else{
            max=Math.max(max,i-current+1);
        }
        map.put(v,i);
    }
    return max;
    }
}