两数之和

给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。

你可以假设每种输入只会对应一个答案。但是，你不能重复利用这个数组中同样的元素。

示例:

给定 nums = [2, 7, 11, 15], target = 9

因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

java实现

方法一：暴力法

public class Solution {
    public static 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[i] + nums[j] == target) {
                   return new int[] {i, j};
                }
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

方法二：两遍哈希表

public class Solution {

    public static 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");
    }
}

方法三：一遍哈希表

public class Solution {

    public static 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");
    }
}

Python3实现

方法一：暴力法

def two_sum(nums, target):
    n = len(nums)
    for i in range(n):
        for j in range(i+1, n):
            if nums[i] + nums[j] == target:
                return [i, j]

方法二：字典模拟Hash

def two_sum(nums, target):
    _dict = {}
    for i, m in enumerate(nums):
        _dict[m] = i
    for i, m in enumerate(nums):
        j = _dict.get(target - m)
        if j is not None and j != i:
            return [i, j]

方法三：一遍字典模拟Hash

def two_sum(nums, target):
    _dict = {}
    for i, m in enumerate(nums):
        if target - m in _dict:
            return [_dict[target - m], i]
        _dict[m] = i