5413. 重新排列句子中的单词_swqjrw

5413. 重新排列句子中的单词

难度中等1

「句子」是一个用空格分隔单词的字符串。给你一个满足下述格式的句子 text :

• 句子的首字母大写
• text 中的每个单词都用单个空格分隔。

请你重新排列 text 中的单词，使所有单词按其长度的升序排列。如果两个单词的长度相同，则保留其在原句子中的相对顺序。

请同样按上述格式返回新的句子。

 

示例 1：

输入：text = "Leetcode is cool"
输出："Is cool leetcode"
解释：句子中共有 3 个单词，长度为 8 的 "Leetcode" ，长度为 2 的 "is" 以及长度为 4 的 "cool" 。
输出需要按单词的长度升序排列，新句子中的第一个单词首字母需要大写。

示例 2：

输入：text = "Keep calm and code on"
输出："On and keep calm code"
解释：输出的排序情况如下：
"On" 2 个字母。
"and" 3 个字母。
"keep" 4 个字母，因为存在长度相同的其他单词，所以它们之间需要保留在原句子中的相对顺序。
"calm" 4 个字母。
"code" 4 个字母。

示例 3：

输入：text = "To be or not to be"
输出："To be or to be not"

 

提示：

• text 以大写字母开头，然后包含若干小写字母以及单词间的单个空格。
• 1 <= text.length <= 10^5

解答：

class Solution {
public:
    
 
static bool comp(string a, string b) {
    return a.size() < b.size();
}
 
void SplitString(const string& s, vector<string>& v, const string& c)
{
    string::size_type pos1, pos2;
    pos2 = s.find(c);
    pos1 = 0;
    
    while (string::npos != pos2)
    {
        v.push_back(s.substr(pos1, pos2 - pos1));
 
        pos1 = pos2 + c.size();
        pos2 = s.find(c, pos1);
    }
    
    if (pos1 != s.length())
        v.push_back(s.substr(pos1));
}
 
string arrangeWords(string text) {
 
    string result;
    vector<string> svec;
    text[0] += 32;
    
    SplitString(text, svec, " ");
    //copy(istream_iterator<string>(text), istream_iterator<string>(), back_inserter(svec));
    //string* arr = new string[svec.size()];
    //copy(svec.begin(), svec.end(), arr);
    
    sort(svec.begin(), svec.end(), comp);
    
    int i = 0, j = svec.size();
    while (i<j)
    {
        result += svec[i];
        if (i + 1 != j)
            result += " ";
        i++;
    }
    
    if (result[0] >= 92 && result[0] <= 122)
        result[0] -= 32;
    return result;
}
};

相同长度的元素，他们的位置发生改变。 

 2.将排序算法改成冒泡排序的方式

#include"stdafx.h"
#include <iostream>
#include<iomanip>
#include<vector>
#include<algorithm>
//#include<unistd.h>
#include<windows.h>
#include<stdlib.h>
#include<time.h>
#include<thread>
 
using namespace std;
 
static bool comp(string a, string b) {
	return a.size() < b.size();
}
 
 
void SplitString(const string& s, vector<string>& v, const string& c)
{
	string::size_type pos1, pos2;
	pos2 = s.find(c);
	pos1 = 0;
 
	while (string::npos != pos2)
	{
		v.push_back(s.substr(pos1, pos2 - pos1));
 
		pos1 = pos2 + c.size();
		pos2 = s.find(c, pos1);
	}
 
	if (pos1 != s.length())
		v.push_back(s.substr(pos1));
}
 
string arrangeWords(string text) {
 
	string result;
	vector<string> svec;
	text[0] += 32;
 
	SplitString(text, svec, " ");
	//copy(istream_iterator<string>(text), istream_iterator<string>(), back_inserter(svec));
	//string* arr = new string[svec.size()];
	//copy(svec.begin(), svec.end(), arr);
 
	//sort(svec.begin(), svec.end(), comp);
 
	//冒泡排序
	 vector<string>::iterator pos1,pos2;
 
 
	for (vector<string>::size_type index = 0; index <= svec.size(); index++)//控制遍数（此处还不完善。因为不一定所有序列都要进行n遍比较）
	{
		pos1=pos2= svec.begin();//初始化让两个迭代器都执行起始点
		++pos2;//总指向比pos1靠前的位置
		while (pos2 != svec.end())//每次循环都进行所有元素的比较
		{
			if ((*pos2).size() < (*pos1).size())
			{
				string tmp = *pos2;
				*pos2 = *pos1;
				*pos1 = tmp;
			}
			++pos1;
			++pos2;
		}
	}
 
	//冒泡排序
 
	int i = 0, j = svec.size();
	while (i<j)
	{
		result += svec[i];
		if (i + 1 != j)
			result += " ";
		i++;
	}
 
	if (result[0] >= 92 && result[0] <= 122)
		result[0] -= 32;
	return result;
}
int main()
{
 
	string text = "Ubnujfo mfurv jphim mzuvlgrjzd ebowhy eptbc xqzjz jtcpdrhpnd lbxdab fxfww qkwohg dvxctqwdx viuvjpreb bsuiolag mqxpmixxij yqpry jzucm gmagayneqw csqhfyjo lgbonmtbgm fplplwoakq qozxlsejr kpeajmxuw esrxcly dbdlfgp isfxag assiub flmprk phoelskteq nntswauaoz cjzkvbz vaflsvubw kfoawx ujmwd ldscnt litcrs wdybdjae iutviyev cdexviqmoq efmhysyap vivch fynyc spcndkpw engjxyg qicxi zvwvspkle jlcgutifu mjdbrgeqd ldqcmod wlzfh xjeqlb wcigq rommz oqwyw saskblv ajfht fqyplksci dvrpelrw ofzdfe pphigymilv wavwrijik fsifiv idsea joedgsz hnxnyjnji eizjbkslu hvmcmiq sknerdxupd cyiyli baymar flabgr llprcfmmzx tvgxyfcj hyxjutckxg ysklwssxph idvexhy pxekhdjc iksjrmbzkt tosfpo dhsichedvv mwxxthr myatzv gdjusdqns gbzwkhvmde roskwgsrdh wufbzvloq kpnzyd fsayvkl ppmskcn rzvoxe axudflwolk mqqjjxp ybnbhgejl llgbljwvc iigugxfdci gscdqig mfhtxmqkz dledk sactj rmmdb gyeglkumtv iasndtrfsq uzkxqo ohuli uqzqxqw eztvy fzutif potijifgc zaywcrmiqz xtpdvh mktee czosln mmmmmoeud robobxpa vxxvxgb vvtvu hnjnprsj 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dfkwcrwjl nfcuh rgzsbsul mzmfomh migzy ydwyfidkvu mghckxkv gacgjr xiqavpc knkijnasm sqrzxpfk zwxlje iplhvae wworcrae ypocojyu zbrcverl wzacqfskca cpmhmwt bodxpaoc aisvcmq qlebkqtbu phnhl lghxf asndqnrkq rzaxewpezm tigqdjd dpgitqpc enrwlqmds ebkhvajh kvbcaa reblxr";
	string result = arrangeWords(text);
 
	char * b = (char*)result.c_str();
	cout << b << endl;
	
	system("pause");
	return 0;
}

报错误：1>e:\letcodepritice\letcode\letcode\letcode.cpp(89): error C2026: 字符串太大，已截断尾部字符

class Solution {
public:
    string arrangeWords(string text) {
        int str_i=0;
        text[0] = text[0]-'A'+'a';
        map<int, vector<string>> mp;
        string st;
        while(text[str_i]!='\0'){
            if(text[str_i]!=' '){
                st.push_back(text[str_i]);
            }
            else{
                if(mp.count((int)st.size())==0){
                    vector<string> v;
                    mp[(int)st.size()] = v;
                }
                mp[(int)st.size()].push_back(st);
                st = "";
            }
            str_i++;
        }
        if(mp.count((int)st.size())==0){
            vector<string> v;
            mp[(int)st.size()] = v;
        }
        mp[(int)st.size()].push_back(st);
        (mp.begin()->second)[0][0] -= ('a'-'A');
        
        string ans;
        for(map<int, vector<string>>::iterator it = mp.begin();it!=mp.end();it++){
            for(int j=0;j<(it->second).size();j++){
                ans.append((it->second)[j]);
                ans.push_back(' ');
            }
        }
        ans.pop_back();
        return ans;
    }
};

参考文献：

https://blog.csdn.net/weixin_41788963/article/details/106175333?fps=1&locationNum=2