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].

解法1，sort後，左標依次移動，右標自適應左移動

與題意不符，題意要求輸出角標，sort後角標變化

 1 public class Solution {
 2     public int[] twoSum(int[] nums, int target) {
 3         Arrays.sort(nums);
 4         int f = 0;
 5         int l = nums.length;
 6         int flag = l - 1;
 7         int[] res = new int[2]; 
 8         for (int
 
 i = 0; i < nums.length; i++) {
 9             
10             while (nums[i] + nums[flag] > target) {
11                 flag--;
12             }
13             if (nums[i] + nums[flag] == target) {
14                 res[0] = i;
15                 res[1] = flag;
16                 return res;
17
 
             }
18         }
19         return res;
20     }
21 }

解法2：hashmap，key存儲數值，value存儲value。

 1 public class Solution {
 2     public int[] twoSum(int[] nums, int target) {
 3         int[] res = new int[2];
 4         Map<Integer, Integer> map = new HashMap<Integer, Integer>();
 5         for (int i = 0; i < nums.length; i++) {
 6             if (map.containsKey(target - nums[i])) {
 7                 res[0] = map.get(target- nums[i]);
 8                 res[1] = i; //邏輯要清晰，res[1] = map.get(nums[i])......

 9 return res; 10 } else { 11  map.put(nums[i], i); 12  } 13  } 14 return res; 15  } 16 }

Two Sum