Question:

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.

Example:

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

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

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

解決：

問題是要從數(shù)組中找到兩個數(shù)據(jù)，使得兩數(shù)之和等于目標值，輸出該兩數(shù)的下標（從0開始）。顯而易見，這里最簡單的是O(n^2)的時間復雜度的解決辦法。

public static int[] twoSum(int[] nums, int target) {
    int[] answer = new int[2];
    A:for (int i = 0; i < nums.length; ++i){
        answer[0] = i;
        int b = target - nums[i];
        for (int j = i + 1; j < nums.length; ++j){
            if (nums[j] == b){
                answer[1] = j;
                break A;
            }
        }
    }
    return answer;
}

考慮O(n)的算法，可以使用map使得查找的復雜度降為O(1)。

public int[] twoSum(int[] nums, int target) {
    int[] answer = new int[2];
    HashMap<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 b = target - nums[i];
        if (map.containsKey(b) && i != map.get(b))
            return new int[]{i, map.get(b)};
     }
     return answer;
}

代碼地址（附測試代碼）：
https://github.com/shichaohao/LeetCodeUsingJava/tree/master/src/TwoSum