LeetCode01 Two Sum

有空还是刷刷leetcode，提神醒脑。

第一题，级别Easy

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.

Example:

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

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

这个题，我用的是O(n^2)的遍历，不知道有没有快速算法。我知道如果给的是排序的数组是有的。

public<span style="font-family: Arial, Helvetica, sans-serif; color: rgb(51, 51, 51); line-height: 30px; widows: 1;"> class TwoSum {</span>

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

for(int i = 0; i< nums.length; i++){
for(int j=i+1; j<nums.length; j++){
if(target == nums[i]+nums[j]){
int[] result = {i,j};
return result;
}
}
}
return null;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] nums = {2, 7, 11, 15};
int target = 9;
int[] result = twoSum(nums, target);
System.out.println("result is "+Arrays.toString(result));
}

}

网上看了看，有O(N)的算法。自己对Map还不是理解恨透啊。可以这样用。

public static int[] twoSumFun2(int[] nums, int target){
Map<Integer, Integer> map = new HashMap<>();
for(int i= 0 ;i<nums.length; i++){
if(map.containsKey(target - nums[i])){
return new int[]{map.get(target-nums[i]), i};
}
map.put(nums[i], i);
}
return null;
}