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  • 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;
         }
     }
    

      

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  • 原文地址:https://www.cnblogs.com/promiseslc/p/8576613.html
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