LeetCode之Easy篇 ——（1）Two Sum

1、Two Sum

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].

解法一：

暴力解决很简单,但时间复杂度为O(n^2)。

 public class Solution {
     public int[] twoSum(int[] nums, int target) {
         int[] result=new int[2];
         for(int i=0;i<nums.length-1;i++){
              for(int j=i+1;j<nums.length;j++){
                 if(nums[i]+nums[j]==target){
                     return new int[]{i,j};
                 }
             }
         }
        return result;
    }
}

解法二：

先遍历一遍数组，建立map数据，然后再遍历一遍，开始查找，找到则记录index。时间复杂度为O(n)。

 public class Solution {
      public int[] twoSum(int[] nums, int target) {
          HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
          int[] res = new int[2];
          for (int i = 0; i < nums.length; ++i) {
              m.put(nums[i], i);
          }
          for (int i = 0; i < nums.length; ++i) {
             int t = target - nums[i];
             if (m.containsKey(t) && m.get(t) != i) {
                 res[0] = i;
                 res[1] = m.get(t);
                 break;
             }
         }
         return res;
     }
 }