Given an array of integers, return indices of the twabout:blanko 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 thesame element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
这是自己第一次做的,暴力解法。时间复杂度:O(n²).
class Solution {
public int[] twoSum(int[] nums, int target) {
int[] a = new int[2];
for(int i = 0; i < nums.length; i++){
for(int j = i + 1;j < nums.length; j++){
if(nums[i] + nums[j] == target){
a[0] = i;
a[1] = j;
}
}
}
return a;
}
}
为了提高时间复杂度,我们需要一个更有效的方法来查找它们的差(即:在数组中查找(target-num[i])的值)。如果它们的差存在,我们就要去查找它的索引。在数组中,最好的查找元素的索引的方法就是HashTable。
HashTable:如果数组中元素的索引位置和元素的值存在对应的关系,这样的数组就称之为哈希表。 公式:index=hash(value);
一般数组索引与元素没有对应关系,而哈希表中索引与元素值存在对应关系,提高了查找数据的效率。
HashTable的具体实现:hashCode方法和equals方法。
方法一:Two-pass Hash Table
时间复杂度:O(n) 空间复杂度:O(n)
public 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 complement = target - nums[i];
if (map.containsKey(complement) && map.get(complement) != i) {
return new int[] { i, map.get(complement) };
}
}
throw new IllegalArgumentException("No two sum solution");
}
方法二:One-pass Hash Table
时间复杂度:O(n) 空间复杂度:O(n)
public int[] twoSum(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");
}
HashMap与Hashtable区别:http://blog.youkuaiyun.com/Ontheroad_/article/details/77367410