1 Two Sum

题目

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].

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解法思路（一）

• 游标 i 在 [0, nums.length) 间遍历，每遍历到一个数，就在其前遍历过的数中找 target - nums[i]，这个找的动作是依赖于哈希表完成的（HashMap），如果找到了，就得到解了，如果没找到，就把当前数放入哈希表，键为 nums[i]，值为 i；

解法实现（一）

时间复杂度

• O(N)；

空间复杂度

• O(N)；

关键词

哈希表 HashMap

package leetcode._1;

import java.util.HashMap;

public class Solution1_1 {

    public int[] twoSum(int[] nums, int target) {

        HashMap<Integer, Integer> toFindComplement = new HashMap<>(nums.length);

        for(int i = 0; i < nums.length; i++) {
            int complement = target - nums[i];
            Integer indices = toFindComplement.get(complement);
            if (indices != null) {
                return new int[]{indices.intValue(), i};
            }
            toFindComplement.put(nums[i], i);
        }

        throw new RuntimeException("No result!");
    }

    public static void main(String[] args) {
        int[] arr = {2, 7, 11, 15};
        int[] result = (new Solution1_1()).twoSum(arr, 9);
        for (int i = 0; i < result.length; i++) {
            if (i == result.length - 1) {
                System.out.println(result[i]);
            } else {
                System.out.print(result[i] + " ");
            }
        }

        System.out.println(result.length);
    }

}