leetcode two sum python 实现

科技前沿 • 2025-02-26 11:47 • 阅读 51

大家好，我是讯享网，很高兴认识大家。

Two Sum-----LeetCode

先百度了一下，看一个解法总感觉有那么些看不懂和复杂.

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

讯享网

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

解题思路：

（1）O(nlogn)。排序，然后两个指针一前一后。因为题中说明了只有一对答案，因此不需要考虑重复的情况。

（2）O(n)。哈希表。将每个数字放在map中，历遍数组，如果出现和数组中的某一个值相加为target的时候，break。这个方法同样适用于多组解的情况。

struct Node { int num, pos; }; bool cmp(Node a, Node b) { return a.num < b.num; } class Solution { public: vector<int> twoSum(vector<int> &numbers, int target) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<int> result; vector<Node> array; for (int i = 0; i < numbers.size(); i++) { Node temp; temp.num = numbers[i]; temp.pos = i; array.push_back(temp); } sort(array.begin(), array.end(), cmp); for (int i = 0, j = array.size() - 1; i != j;) { int sum = array[i].num + array[j].num; if (sum == target) { if (array[i].pos < array[j].pos) { result.push_back(array[i].pos + 1); result.push_back(array[j].pos + 1); } else { result.push_back(array[j].pos + 1); result.push_back(array[i].pos + 1); } break; } else if (sum < target) { i++; } else if (sum > target) { j--; } } return result; } };

讯享网
懒的建工程，就用python写了一个

讯享网def twosum(array, target): alen = len(array) for i in range(alen): for j in range(i + 1, alen): if (array[i] + array[j] == target): return i, j return 0, 0 arr = [2, 7, 11, 15, 1, 3, 53, 245] print twosum(arr, 56)

leetcode two sum python 实现

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