每日算法练习（2020-1-17）

Posted 2021-03-15 qyx66

在此祝各位同仁，小年快乐!

 

 

package com.qyx.Tree;
import java.util.ArrayList;
import java.util.List;
/**
 * 
 * @author Q
 *
 */
public class PhoneNumber {
/*使用二叉树完成题解*/
     private String letterMap[] = {
                " ",    //0
                "",     //1
                "abc",  //2
                "def",  //3
                "ghi",  //4
                "jkl",  //5
                "mno",  //6
                "pqrs", //7
                "tuv",  //8
                "wxyz"  //9
        };
     private ArrayList<String> res;
     private void findCombination(String digits,int index,String s)
     {
        if(index==digits.length())
        {
            /*长度和电话号码长度相等直接加入list，退出递归*/
            res.add(s);
            return;
        }
        Character c=digits.charAt(index);
        /*寻找数字对应的字母们*/
        String letters=letterMap[c-‘0‘];
        for(int i=0;i<letters.length();i++){
            findCombination(digits, index+1, s+letters.charAt(i));
        }
        return;
     }
     public List<String> letterCombinations(String digits)
     {
         res=new ArrayList<>();
         if(digits.equals(""))
         {
             return res;
         }
         findCombination(digits, 0, "");
         return res;
     }
}

 

 

package com.qyx.Tree;
import java.util.Arrays;
/**
 * 
 * @author Q
 *
 */
public class ThreeNumberSum {
    /*使用双指针*/
    public int threeSumClosest(int[] nums,int target)
    {
        //对数组进行排序，花费时间复杂度O(logN)
        Arrays.sort(nums);
        int ans=nums[0]+nums[1]+nums[2];
        /*花费时间复杂度O(N^2)*/
        for(int i=0;i<nums.length;i++)
        {
            /*头指针在i+1处*/
            int start=i+1;
            /*尾指针在nums.length-1*/
            int end=nums.length-1;
            while(start<end)
            {
                
                int sum=nums[i]+nums[start]+nums[end];
                if(Math.abs(target-sum)<Math.abs(target-ans))
                {
                    //如果sum的距离里target则更新ans的值
                    ans=sum;
                }else if(sum<target)
                {
                    start++;
                }else if(sum>target)
                {
                    end--;
                }else{
                    return ans;
                }
            }
        }
        return ans;
    }

}