[LeetCode] 1. Two Sum 两数和

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].

给一个没排序的包含整数的数组，找出使得两个数的和是给定值的indices。假设每个输入只有一个答案， 同一元素不会用到两次。

解法：

1. 暴力解法，两个for循环遍历，两个数相加和目标数比较。Time: O(n^2)

2. 先对数组快速排序，然后用两个指针分别指向头和尾，每次比较头尾两个数的和，如果比target小，头标记右移，如果大，尾标记左移。需要注意记录快速排序前后数字的位置变化。Time: O(n)

3. 先遍历一遍数组，建立数字和index的HashMap，然后再遍历一遍，开始查找target - num[i]是否在map中，如果在，找到并返回index。Time: O(n)  Space: O(n)

Java：解法1

public static int[] twoSum(int[] numbers, int target) {
    int[] ret = new int[2];
    for (int i = 0; i < numbers.length; i++) {
        for (int j = i + 1; j < numbers.length; j++) {
            if (numbers[i] + numbers[j] == target) {
                ret[0] = i + 1;
                ret[1] = j + 1;
            }
        }
    }
    return ret;
}　　

Java：解法2

class Pair implements Comparable<Pair>{
    public int number;
    public int idx;
    public Pair(int number, int idx){
        this.number = number;
        this.idx = idx;
    }
    public int compareTo(Pair other){
        return this.number - other.number;
    }
}
public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int n = numbers.length;
        Pair[] pairs = new Pair[n];
        for(int i = 0; i < n; ++i){
            pairs[i] = new Pair(numbers[i], i + 1);
        }
        Arrays.sort(pairs);
        int [] result = new int[2];
        int begin = 0;
        int end = n - 1;
        while(begin < end){
            if(pairs[begin].number + pairs[end].number < target){
                begin++;
            }
            else if (pairs[begin].number + pairs[end].number > target){
                end--;
            }
            else{
                if(pairs[begin].idx > pairs[end].idx){
                    result[0] = pairs[end].idx;
                    result[1] = pairs[begin].idx;
                }else{
                result[0] = pairs[begin].idx;
                result[1] = pairs[end].idx;
                }
                break;
            }
        }
        return result;
    }
}　　

Java: 解法3，Two loops

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
        int[] res = new int[2];
        for (int i = 0; i < nums.length; ++i) {
            m.put(nums[i], i);
        }
        for (int i = 0; i < nums.length; ++i) {
            int t = target - nums[i];
            if (m.containsKey(t) && m.get(t) != i) {
                res[0] = i;
                res[1] = m.get(t);
                break;
            }
        }
        return res;
    }
}

Java: 解法3，one loop

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
        int[] res = new int[2];
        for (int i = 0; i < nums.length; ++i) {
            if (m.containsKey(target - nums[i])) {
                res[0] = i;
                res[1] = m.get(target - nums[i]);
                break;
            }
            m.put(nums[i], i);
        }
        return res;
    }
}

Python:

class Solution(object):
    def twoSum(self, nums, target):
        hash_map = {}
        for i, v in enumerate(nums):
            hash_map[v] = i

        for index1, value in enumerate(nums):
            if  target - value in hash_map:
                index2 = hash_map[target - value]
                if index1 != index2:
                    return [index1, index2]　　

Python:

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        lookup = {}
        for i, num in enumerate(nums):
            if target - num in lookup:
                return [lookup[target - num], i]
            lookup[num] = i

Python: wo

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        lookup = {}
        for i in range(len(nums)):
            if target - nums[i] in lookup:
                return [lookup[target - nums[i]], i]
            lookup[nums[i]] = i　　

Python:

class Solution(object):
    def twoSum(self, nums, target):
        if len(nums) <= 1:
            return False
        buff_dict = {}
        for i in range(len(nums)):
            if nums[i] in buff_dict:
                return [buff_dict[nums[i]], i]
            else:
                buff_dict[target - nums[i]] = i

C++:

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> lookup;
        for (int i = 0; i < nums.size(); ++i) {
            if (lookup.count(target - nums[i])) {
                return {lookup[target - nums[i]], i};
            }
            lookup[nums[i]] = i;
        }
        return {};
    }
};

　　

　　

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[LeetCode] 167. Two Sum II - Input array is sorted 两数和 II - 输入是有序的数组　

[LeetCode] 170. Two Sum III - Data structure design 两数之和之三 - 数据结构设计　

[LeetCode] 653. Two Sum IV - Input is a BST 两数之和之四 - 输入是二叉搜索树

　　

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