1. Two Sum

Description:

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].

My code:

public class Solution {
  public int[] twoSum(int[] nums, int target) {
    int[] ans = 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){
          ans[0] = i;
          ans[1] = j;
          break;
        }
    }
  }
  return ans;
  }
}


问题:代码复杂度过高,抽空看一下复杂度为O(n)的代码。
posted @ 2017-02-27 20:58  程序媛家  阅读(124)  评论(0编辑  收藏  举报