Leetcode TwoSum & 两数之和解题报告

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这篇博客介绍了如何使用哈希表来解决在给定数组中找到两个数，使得它们的和为目标值的问题。提供了两种Java实现方式：一种是双哈希表遍历，另一种是单次哈希表遍历。这两种方法都确保了唯一解且不使用重复元素。示例代码展示了具体的实现细节和逻辑。

Two Sum

给定一个数组nums和一个整数值target，返回两个数值（nums数组下标的index），使其所对应的数组元素，相加得到目标值target。

附加条件约束：

每个nums数组和目标值，只有一个解决方案
不能使用重复的值作为解，eg: 列子 Example 2中目标值是6，不能直接返回 [0,0]
可以按照任何顺序返回答案

Describe

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have *exactly* one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

2 <= nums.length <= 103
-109 <= nums[i] <= 109
-109 <= target <= 109
Only one valid answer exists.

java

//version 1 两次hash table ； 可读性相对高
public static int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            map.put(nums[i], i);
        }
        for (int i = 0; i < nums.length; i++) {
            int searchWithMap = target - nums[i];
            if (map.containsKey(searchWithMap) && i != map.get(searchWithMap)) {
                return new int[]{i, map.get(searchWithMap)};
            }
        }
        return null;
}



 //version 2 单次hash table； 可读性相对低
 public static int[] twoSumOneHashTable(int[] nums, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            int complement = target - nums[i];
            if (map.containsKey(complement)) {
                return new int[] { map.get(complement), i };
            }
            map.put(nums[i], i);
        }
        throw new IllegalArgumentException("No two sum solution");
    }

python


def two_sum(nums, target):
    ele_map = {}
    result = [-1, -1]
    for index, ele in enumerate(nums):
        ele_map[ele] = index
    for index, ele in enumerate(nums):
        if target - ele in ele_map and index != ele_map.get(target - ele):
            result[0] = index
            result[1] = ele_map.get(target - ele)
    return result