算法基础-字符移位

	用STL的partition算法直接搞定



	



	#include<iostream>



	#include<algorithm>



	using namespace std;



	 



	bool isLower(charc) {



	    returnc>='a';



	}



	 



	intmain() {



	    string s;



	    while(cin >> s) {



	        stable_partition(s.begin(),s.end(),isLower);



	        cout << s << endl;



	    }



	}

#短小精悍
def fun(l):
    l = list(l)
    i = 0
    count = 0
    while count < len(l):
        if l[i].isupper():
            l.append(l.pop(i))
        else:
            i += 1
        count += 1
    return reduce(lambda x,y: x+y,l)

	import sys



	import re  



	for line in sys.stdin:



	    s1 = line.strip() 



	    print(re.sub(r'[A-Z]','',s1) + re.sub(r'[a-z]','',s1))

#include<iostream>
#include<string>
using namespace std;
int main()
{string str;
while(cin>>str)
{
int len=str.size();
for(int i=0,count=0;i<len,count<len;i++,count++)
{
if(str[i]>='A'&&str[i]<='Z')
{
char w=str[i];
for(int j=i+1;j<len;j++)
str[j-1]=str[j];
str[len-1]=w;
i=i-1;
}
}
cout<<str<<endl;
}
}

import java.util.Scanner;

/**
 * Created by 10648 on 2017/2/22 0022.
 */
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        while (scanner.hasNext()) {
            String string = scanner.nextLine();
            char [] chars = string.toCharArray();
            int length = string.length();
            for (int i = 0; i < length; i ++) {
                for (int j = 1; j < length - i; j ++) {
                    if ((chars[j-1] >= 'A' && chars[j-1] <= 'Z') &&
                            (chars[j] < 'A' || chars[j] > 'Z')) {
                        char temp = chars[j-1];
                        chars[j-1] = chars[j];
                        chars[j] = temp;
                    }
                }
            }
            System.out.println(String.valueOf(chars));
        }
    }
}

#include <iostream>
#include <string>
using namespace std;
 
int main() {
    string s;
    while (cin >> s) {
        int n = s.size();
        int p = n;
        for (int i = 0;i < p;++i) {
            if (s[i] >= 'A' && s[i] <= 'Z') {
                char t = s[i];
                int j = i;
                for (;j < n-1;++j) {
                    s[j] = s[j + 1];
                }
                s[j] = t;
                --p;
                --i;
            }
        }
        cout << s << endl;
    }
 
    return 0;
}

#include<iostream>
#include<string>
#include<algorithm> 
using namespace std;
int main()
{
	string	A;
	while(cin>>A)
	{
		int length = A.size();
		for(int i = length-2;i>=0;i--)
		{
		for(int j = i;(i<length) &&( (A[j]>='A'&&A[j]<='Z') && 
                                         (A[j+1]>='a' && A[j+1]<='z'));j++)
				     swap(A[j],A[j+1]);
		}
		cout<<A<<endl;		
	}
	return 0;
}

#include <iostream>
#include <string.h>
using namespace std;
int main(){
	string s;
	while(cin >> s){
		if(s.length() >= 1 && s.length() <= 1000){
			for(int i = 0; i < s.length(); i++)
				if(s[i] >= 'a' && s[i] <= 'z')
					cout << s[i];
			for(int i = 0; i < s.length(); i++)
				if(s[i] <= 'Z' && s[i] >= 'A')
					cout << s[i];
			cout << endl;//开始没写这个老是通不过...
		}
	}
	return 0;
}
根据题意输出了正确结果...
这个题仔细看看答案- -把大小写分开输出就可以了。
第一次做对腾讯的题，好鸡冻2333.

C++模拟

类似冒泡排序

• 若是大写字母，不进行操作
• 若是小写字母，则将该小写字母移动到其之前的所有大写字母之前

#include <bits/stdc++.h>

using namespace std;

int main() {
    string s;
    while (cin >> s) {
        int n = s.size();
        for (int i = 1; i < n; ++i) {
            if (isupper(s[i])) continue;    // 大写字母，不进行操作
            for (int j = i - 1; j >= 0; --j) {    // 小写字母，移动到其之前的所有大写字母之前
                if (isupper(s[j])) swap(s[j], s[j + 1]);
            }
        }
        cout << s << endl;
    }
    return 0;
}

    public  String getResult(String str){
        char[] c = str.toCharArray();
        int index = 0;
        int big = 0;
        while (big<c.length&&index<c.length){
            if (isSmall(c[index])){        //说明碰到小写字母了
                if(big<index){  //说明之前有big出现
                    if (isSmall(c[big+1])){ //big的下一个是小写字母 那么我们可以直接交换
                        char temp = c[big];
                        c[big] = c[index];
                        c[index] = temp;
                        big = index;
                    }else {     //big的下一个还是big 如果贸然交换 相对顺序就变了 所以我们要插入排序
                        char value = c[index];
                        while (index>big){
                            c[index] = c[index-1];
                            index--;
                        }
                        c[index] = value;
                        big = big+1;
                    }
                }else { //这种情况说明之前还没有big出现
                    big++;
                }
            }
            index++;
        }
        return String.valueOf(c);
    }