題目描述

Two Sum（兩數(shù)之和）

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].

分析

咋一看這道題就知道用暴力迭代肯定能解決問(wèn)題，但是LeetCode肯定不會(huì)接受這種暴力搜索這么簡(jiǎn)單的方法（其時(shí)間復(fù)雜度$O(n^2)$）。

解法1

先遍歷一遍數(shù)組，建立map<key=(數(shù)組值), vale=(索引)>數(shù)據(jù)。然后再遍歷一遍，開(kāi)始超找，找到則記錄index。代碼如下：

C++

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

Swift

class Solution {
    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var map: [Int: Int] = [:]
        for i in 0..<nums.count {
            map[nums[i]] = i
        }
        var res = [Int]()
        for i in 0..<nums.count {
            let t = target - nums[i]
            if let index = map[t], map[t] != i {
                res.append(i)
                res.append(index)
                break;
            }
        }
        return res
    }
}

解法2

或者我們可以寫的更加簡(jiǎn)潔一些，把兩個(gè)for循環(huán)合并成一個(gè)：

C++

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

Swift

class Solution {
    func twoSum(_ nums: [Int], _ target: Int) -> [Int] {
        var map: [Int: Int] = [:]
        for i in 0..<nums.count {
            if let index = map[target - nums[i]] {
                return [index, i]
            }
            map[nums[i]] = i
        }
        return []
    }
}