#1 Tow Sum

My solution:

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        for(int i=0;i<nums.length-1;i++){
            for(int j=i+1;j<nums.length;j++){
                if(nums[i]+nums[j]==target){
                    result[0]=i;
                    result[1]=j;
                    return result;
                }
            }
        }
        return result;
    }
}

reference solution:

Approach 1: Brute force

time complexity: O(n2)                  

space complexity: O(1)

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

对比后需要改进的地方： 

1.  没有必要新声明一个result，直接返回更简洁

return new int[] {i, j};

2. 没有找到结果时，不应该返回默认result赋值，应该学会抛出异常

throw new IllegalArgumentException("No tow sum solution");

 

Approach 2:Two-pass Hash Table

What is the best way to maintain a mapping of each element in the array to its index? A hash table.

time complexity: O(n)                  

space complexity: O(n)

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        map.put(nums[i], i);
    }
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement) && map.get(complement) != i) {
            return new int[] { i, map.get(complement) };
        }
    }
    throw new IllegalArgumentException("No two sum solution");
}

注意点：

1. map的声明

Map<Integer, Integer> map = new HashMap<>();

2. map的方法   (关于hashmap：https://www.cnblogs.com/skywang12345/p/3310835.html#b2

map.put(nums[i], i)
map.containsKey(complement)
map.get(complement)

 

Approach 3: One-pass Hash Table

While we iterate and inserting elements into the table, we also look back to check if current element's complement already exists in the table.

time complexity: O(n)                  

space complexity: O(n)

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement)) {
            return new int[] { map.get(complement), i };
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution");
}