算法leetcode-1.Two Sum

原创于 2018-06-05 21:49:40 发布 · 342 阅读

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本文介绍了一种解决两数之和问题的有效算法。通过使用哈希表存储目标值减去当前元素值得到的差值及其索引，可以快速找到满足条件的两个数。这种方法时间复杂度为O(n)，避免了传统双重循环的低效。

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:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        Map<Integer, Integer>  numMap = new HashMap<>();
        for (int i = 0; i < numMap.length; i++) {
            int num = nums[i];
            if (numMap.containsKey(num)) {
                return new int[] {numMap.get(num), i};
            }
         numMap.put(target - num, i);
        } 
        return null; 
    }
}