劍指offer java實現合集（7）第31~35題

31.整數中1出現的個數

相當於計算一個數中出現的個數，依次相加即可。

public class Solution {
    public int NumberOf1Between1AndN_Solution(int n) {
        int count = 0;
        for(int i =1;i<=n;i++){
            int num = getnumber(i);
            count+=num;
        }
        return count;
    }
    public int getnumber(int n){
        int count = 0;
        while(n!=0){
            int num = n%10;
            if(num==1){
                count++;
            }
            n=n/10;
        }
        return count;
    }
}

32.把數組排成最小的數

這是我在評論區看到大佬的思路，定義了一個比較規則

import java.util.ArrayList;
import java.util.*;
public class Solution {
    public String PrintMinNumber(int [] numbers) {
        int len = numbers.length;
        if(len==0){
            return "";
        }
        String []str = new String[len];
        StringBuilder sb = new StringBuilder();
        for(int i = 0;i<len;i++){
            str[i] = String.valueOf(numbers[i]);
        }
        Arrays.sort(str,new Comparator<String>(){
            public int compare(String s1,String s2){
                String c1 = s1+s2;
                String c2 = s2+s1;
                return c1.compareTo(c2);
            }
        });
        for(int i =0;i<len;i++){
            sb.append(str[i]);
        }
        return sb.toString();
    }
}

33.醜數

 

import java.util.*;
public class Solution {
    public int GetUglyNumber_Solution(int index) {
        ArrayList<Integer> arr = new ArrayList();
        if(index==0){
            return 0;
        }
        arr.add(1);
        int i = 1;
        int index2 = 0;
        int index3 = 0;
        int index5 = 0;
        while(i<index){
            int num2 = arr.get(index2)*2;
            int num3 = arr.get(index3)*3;
            int num5 = arr.get(index5)*5;
            int min = Math.min(num2,Math.min(num3,num5));
            arr.add(min);
            i++;
            if(min==num2){
                index2++;
            }
            if(min==num3){
                index3++;
            }
            if(min==num5){
                index5++;
            }
        }
        return arr.get(arr.size()-1);
    }
}

34.第一個只出現一次的字符

借用了hashmap存儲了每個字符的出現次數，再遍歷整個數組，判斷每一位在hashmap中存儲的數量值。當找到只出現一次的字母時，退出循環，返回結果。

import java.util.*;
public class Solution {
    public int FirstNotRepeatingChar(String str) {
        HashMap<Character,Integer> hash = new HashMap();
        char[] cha = str.toCharArray();
        for(char x:cha){
            if(hash.containsKey(x)){
                hash.put(x,2);
            }else{
            hash.put(x,1);
            }
        }
        int res = -1;
        for(int i =0;i<str.length();i++){
            if(hash.get(cha[i])==1){
                res=i;
                break;
            }
        }
        return res;
    }
}

35.數組中的逆序對

這是十分不推薦的暴力解法，牛客網只能通過50%的用例，會超時。

正確的應該是 用歸併排序的思想，先挖個坑，過幾天回來填。

public class Solution {
    public int InversePairs(int [] array) {
        int count = 0;
        for(int i =0;i<array.length-1;i++){
            for(int j =i+1;j<array.length;j++){
                if(array[i]>array[j]){
                    count++;
                }
            }
        }
        return count%1000000007;
    }
}