Two Sum

题目

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:

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Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

思路

可以建一个map，存储遍历过的数。当遇到新的数字时，判断map里头有没有 target 与该数字的差。

解答

Cpp

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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 {};
    }
};

Python

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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)
            else:
                lookup[nums[i]] = i

JavaScript

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/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number[]}
 */
var twoSum = function(nums, target) {
    let lookup = new Map();
    for (let i=0; i<nums.length; ++i) {
        if (lookup.has(target - nums[i])) {
            return [lookup.get(target-nums[i]), i];
        } else {
            lookup.set(nums[i], i);
        }
    }
};

复杂度分析

  • 时间复杂度：\(O(n)\)
  • 空间复杂度：\(O(n)\)

Comments

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