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].
idea
本题和lintcode 题目有点区别,本题允许两个重复数出现,比如数组 [3, 3] target = 6,如果按照 lintcode 代码会出现,results = [1, 1] 而实际上我们要的是 [0, 1],所以需要用
if (hash.containsKey(complement) && hash.get(complement) != i)
做一个特判,保证在给 results[0] 和 results[1] 赋值时不出现重复,但还有一点需要强调的是之前在赋值时写的代码是results[1] = hash.get(nums[i])
,这么写在出现上述 [3, 3] 这种带有元素的重复数组会有问题,尽管 if 条件句做了特判,但赋值时可能hash.get(complement)
和hash.get(nums[i])
得到的是同一个值,所以要直接写results[1] = i
Code
- 遍历两次 hashTable
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] results = new int[2];
if (nums == null || nums.length == 0) {
return results;
}
Map<Integer, Integer> hash = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
hash.put(nums[i], i);
}
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (hash.containsKey(complement) && hash.get(complement) != i) {
results[0] = hash.get(complement);
results[1] = i;
}
}
return results;
}
}
- 遍历一次 hashTable
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] results = new int[2];
if (nums == null || nums.length == 0) {
return results;
}
Map<Integer, Integer> hash = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int complement = target - nums[i];
if (hash.containsKey(complement) && hash.get(complement) != i) {
results[0] = hash.get(complement);
results[1] = i;
}
hash.put(nums[i], i);
}
return results;
}
}