<如何在 LeetCode 上解决「两数之和」问题？ - 从算法思维到编程实现>

前端

2022-12-28 03:36:16

「两数之和」的算法奥秘：从概念到实践

算法思维：庖丁解牛

当我们面对「两数之和」这一经典算法问题时，与其仓促求解，不如先抽丝剥茧，层层深入地理解其算法思维。问题的本质在于，我们需要在给定数组中找出两数之和等于指定目标值的索引对。为此，我们可以将其分解为以下几个步骤：

遍历数组： 从第一个元素开始，依次遍历整个数组中的每一个元素。
比较元素： 对于每个元素，将其与之前遍历过的所有元素进行比较，检查是否存在一对元素之和等于指定的目标值。
存储索引： 如果找到满足条件的元素对，将它们的索引存储起来。

编程实现：代码千行，殊途同归

理解了算法思维后，我们就可以着手编写代码了。虽然不同的编程语言语法各异，但算法的核心思想却是一致的。以下是用 Python、Java、C++ 和 JavaScript 实现「两数之和」算法的代码示例：

Python

def two_sum(nums, target):
    """
    Finds the indices of two numbers in a list that sum to a target value.

    Args:
    nums: A list of numbers.
    target: The target value to find the sum of.

    Returns:
    A list of two indices that sum to the target value, or an empty list if no such pair exists.
    """

    # Create a dictionary to store the indices of the numbers we have seen so far.
    seen = {}

    # Iterate over the list of numbers.
    for i, num in enumerate(nums):
        # Calculate the complement of the target value.
        complement = target - num

        # Check if the complement is in the dictionary.
        if complement in seen:
            # Return the indices of the two numbers that sum to the target value.
            return [seen[complement], i]

        # Add the current number and its index to the dictionary.
        seen[num] = i

    # No pair of numbers sums to the target value.
    return []

Java

public class TwoSum {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> seen = new HashMap<>();

        for (int i = 0; i < nums.length; i++) {
            int complement = target - nums[i];

            if (seen.containsKey(complement)) {
                return new int[] { seen.get(complement), i };
            }

            seen.put(nums[i], i);
        }

        return new int[0];
    }
}

C++

vector<int> twoSum(vector<int>& nums, int target) {
  unordered_map<int, int> seen;

  for (int i = 0; i < nums.size(); i++) {
    int complement = target - nums[i];

    if (seen.find(complement) != seen.end()) {
      return {seen[complement], i};
    }

    seen[nums[i]] = i;
  }

  return {};
}

JavaScript

const twoSum = (nums, target) => {
  const seen = {};

  for (let i = 0; i < nums.length; i++) {
    const complement = target - nums[i];

    if (seen[complement] !== undefined) {
      return [seen[complement], i];
    }

    seen[nums[i]] = i;
  }

  return [];
};