我在尝试在向量中找到多个相同元素时遇到问题。我想每3个索引找到重复的元素。如果有重复,则接下来的2个元素将进入新的向量。
std::vector<std::wstring> wsBuf = {"one" , "three", "five", "two", "one","ten", "nine"};`
std::vector<std::wstring> wsWriteFile;
for(DWORD i = 0; i < wsBuf.size(); ++i)
{
std::vector<std::wstring>::iterator it = wsReadFile.begin(); //wsReadFile has an enumerated list
it = std::find(it, wsReadFile.end(), wsBuf[i]);
if(it != wsReadFile.end())
{
DWORD index = std::distance(wsReadFile.begin(), it);
if(index < wsReadFile.size())
{
wsWriteFile.push_back(wsReadFile[index];
wsWriteFile.push_back(wsReadFile[index + 1];
wsWriteFile.push_back(wsReadFile[index + 2];
}
}
}
这只会找到零指数,但不能找到第二个指数。有没有人知道如何用C ++ 11函数找到第二个索引?
我找到了一种没有c ++ 11函数的方法
`for(DWORD i = 0; i < wsBuf.size(); ++i)
{
DWORD j = 0;
do
{
if(wsBuf[i] == wsReadFile[j])
{
wsWriteFile.push_back(wsReadFile[index];
wsWriteFile.push_back(wsReadFile[index + 1];
wsWriteFile.push_back(wsReadFile[index + 2];
}
j+=3;
}while(j < wsReadFile.size());
}`
答案 0 :(得分:0)
对于每个wsBuf
,您只需拨打find
一次。这清楚地找到......一个元素。
it = std::find(it, wsReadFile.end(), wsBuf[i]);
if(it != wsReadFile.end())
{
DWORD index = std::distance(wsReadFile.begin(), it);
if(index < wsReadFile.size())
{
wsWriteFile.push_back(wsReadFile[index];
wsWriteFile.push_back(wsReadFile[index + 1];
wsWriteFile.push_back(wsReadFile[index + 2];
}
}
它也没有编译,你遗漏了一些)
。将来,编写并演示问题的邮政编码;如果你不知道这个问题,你就无法知道解决它的问题。
无论如何,你需要一个循环。我也想要边界检查。并且减去迭代器&lt; - &gt;索引转换。更少的类型无关紧要。
auto it = std::find(wsReadFile.begin(), wsReadFile.end(), wsBuf[i]);
while(it != wsReadFile.end())
{
auto left = std::distance(it, wsReadFile.end());
// don't copy past the end:
auto copy_end = it+(std::min)(left, decltype(left)(3));
std::copy( it, copy_end, std::back_inserter(wsWriteFile) );
// find next:
it = std::find( copy_end, wsReadFile.end(), wsBuf[i]);
}