LeetCode第一题—— Two Sum（寻找两数，要求和为target）

题目描述:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

My solution（50ms,38.5MB）

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

以下是标准答案：

Approach 1: Brute Force（16ms,38.5MB）

class Solution {
    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");
   }
}

Approach 2: Two-pass Hash Table（2ms,38.1MB）

class Solution {
      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");
      }
}

Approach 3: One-pass Hash Table（2ms,38.2MB）

class Solution {
    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)) {//判断键名是否包含complement
                return new int[] { map.get(complement), i };
            }
            map.put(nums[i], i);
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}