1. Sum of two numbers

Solutions 1

Violence law solves problems , Double loop to find that the sum of two numbers in the array is target The number of .

    public int[] twoSum(int[] nums, int target) {
    
        int n = nums.length;
        for (int i = 0; i < n-1; i++) {
    
            for (int j = i+1; j < n; j++) {
    
                if(nums[i]+nums[j]==target)
                    return new int[]{
    i,j};
            }
        }
        throw new IllegalArgumentException("no solution");
    }

Solutions 2

Use hash Tabular Key-value structure , Save the numbers —> Index relationship . According to the target number, come to hash Find in table . Deposit first hash surface , And then query .

    //2. Use Hash surface 
    public int[] twoSum1(int[] nums, int target) {
    
        // newly build hashmap
        HashMap<Integer, Integer> map = new HashMap<>();
        //1. Save all data to map in 
        for (int i = 0; i < nums.length; i++) {
    
            map.put(nums[i],i);
        }
        // Traversal array again , Find the number corresponding to each number 
        for (int i = 0; i < nums.length; i++) {
    
            int targetNum = target-nums[i];
            if(map.containsKey(targetNum)&& map.get(targetNum)!=i)
                return new int[]{
    i,map.get(targetNum)};
        }
        throw new IllegalArgumentException("no solution");
    }

Solutions 3

Edge save to hash In the table , While searching in the table .

    //3. The improved hashmap
    public int[] twoSum2(int[] nums, int target) {
    
        // newly build hashmap
        HashMap<Integer, Integer> map = new HashMap<>();
        //1. Save all data to map in 
        for (int i = 0; i < nums.length; i++) {
    
            int targetNum = target-nums[i];
            if(map.containsKey(targetNum)&& map.get(targetNum)!=i)
                return new int[]{
    map.get(targetNum),i};
            map.put(nums[i],i);
        }
        throw new IllegalArgumentException("no solution");
    }