56. Two Sum

描述

给一个整数数组，找到两个数使得他们的和等于一个给定的数 target。
你需要实现的函数twoSum需要返回这两个数的下标, 并且第一个下标小于第二个下标。注意这里下标的范围是 0 到 n-1。

注意事项

你可以假设只有一组答案

样例

给出 numbers = [2, 7, 11, 15], target = 9, 返回 [1, 2]

挑战

Either of the following solutions are acceptable:
O(n) Space, O(nlogn) Time
O(n) Space, O(n) Time

代码

1. hash 遍历 hash 表一次，时间复杂度 O(n)，空间复杂度 O(n)
   hash 表里键是数值的大小，值是数组的下标

public class Solution {
    /*
     * @param numbers: An array of Integer
     * @param target: target = numbers[index1] + numbers[index2]
     * @return: [index1 + 1, index2 + 1] (index1 < index2)
     */
    public int[] twoSum(int[] numbers, int target) {
        Map hash = new HashMap<>();
        int[] results = new int[2];
        
        for (int i = 0; i < numbers.length; i++) {
            int complement = target - numbers[i];
            if (hash.containsKey(complement)) {
                // 注意顺序，complement 早就加入了 hash，所以 complement 的 Index 比 i 小
                // 下标从 1 开始
                results[0] = hash.get(complement);
                results[1] = i;
                return results;
            }
            hash.put(numbers[i], i);
        }
        
        return results;
    }
}

2. 暴力解法两重for循环 时间复杂度O(n^2)，空间复杂度O(1)

public class Solution {
    /*
     * @param numbers: An array of Integer
     * @param target: target = numbers[index1] + numbers[index2]
     * @return: [index1 + 1, index2 + 1] (index1 < index2)
     */
    public int[] twoSum(int[] nums, int target) {
        Map hash = new HashMap<>();
        int[] results = 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) {
                    results[0] = i;
                    results[1] = j;
                    return results;
                }
            }
        }
        return results;
    }
}

3. hash 遍历 hash 表两次，时间复杂度 O(n)，空间复杂度 O(n)

public class Solution {
    /*
     * @param numbers: An array of Integer
     * @param target: target = numbers[index1] + numbers[index2]
     * @return: [index1 + 1, index2 + 1] (index1 < index2)
     */
    public int[] twoSum(int[] nums, int target) {
        Map hash = new HashMap<>();
        int[] results = new int[2];
        
        for (int i = 0; i < nums.length; i++) {
            hash.put(nums[i], i);
        }
        
        for (int i = 0; i < nums.length; i++) {
            int complement = target - nums[i];
            if (hash.containsKey(complement)) {
                results[0] = i + 1;
                results[1] = hash.get(complement) + 1;
                return results;
            }
        }
        
        return results;
    }
}