[Leetcode] 1. Two Sum-优快云博客

本文链接：https://blog.youkuaiyun.com/Renoler/article/details/102267056

本文提供了三种解决LeetCode经典题目“两数之和”的算法方案：暴力破解法，两次遍历哈希表法及一次遍历哈希表法。通过对比不同算法的优劣，帮助读者理解哈希表在提升查找效率中的应用。

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Problem: https://leetcode.com/problems/two-sum/solution/

Solution1:
Brute Force

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        for (int i = 0; i < nums.length-1; i++) {
            for (int j = i+1; j < nums.length; j++) {
                if ((nums[i] + nums[j]) == target) {
                    result[0] = i;
                    result[1] = j;
                    break;
                }
            }
            if (result[0] != result[1]) {
                break;
            }
        }
        return result;
    }
}

Solution2:
遍历两次hashmap：第一次存数，第二次取数

class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
            map.put(nums[i], i); // 将数字和对应的index存入hashmap
        }
        for(int i = 0; i < nums.length; i++) {
            int diff = target-nums[i];// 寻找满足target的另一半是否已经存在
            						  // 需要注意的是这个另一半的index不可以是本身
            if (map.containsKey(diff) && map.get(diff) != i) {
                return new int[]{i, map.get(diff)};
            }
        }
        return null;
    }
}

Solution3:
遍历一次hashmap：在put之前现寻找是否满足target的另一半

class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
            if(map.containsKey(target-nums[i])) {
                return new int[] {i, map.get(target-nums[i])}; //先寻找，再插入
            }
            map.put(nums[i], i);
        }
        return null;
    }
}