Java 如何从两组具有备用值的列表中生成字符串
我有两个列表,其中包含一些值,我必须从中生成字符串,这样我将获取第一个列表的第一个值和第二个列表的第一个值,以及第一个列表的第二个值和第二个列表的第二个值,依此类推。假设这两个列表包含面试时间。所以我在这里给出我的代码Java 如何从两组具有备用值的列表中生成字符串,java,Java,我有两个列表,其中包含一些值,我必须从中生成字符串,这样我将获取第一个列表的第一个值和第二个列表的第一个值,以及第一个列表的第二个值和第二个列表的第二个值,依此类推。假设这两个列表包含面试时间。所以我在这里给出我的代码 List<String> interviewTimingToFrom1 = Arrays.asList(interviewTime1.split(",")); for(String a :interviewTimingToFrom1){ System.out.
List<String> interviewTimingToFrom1 = Arrays.asList(interviewTime1.split(","));
for(String a :interviewTimingToFrom1){
System.out.println("Timing 1:"+a);
}
List<String> interviewTimingToFrom2 = Arrays.asList(interviewTime2.split(","));
for(String a :interviewTimingToFrom2){
}
所以现在我需要制作一个字符串,比如从12.00am到1.00am,从2.00am到3.00am,我是如何做到的。请帮助int maxSize=Math.max(interviewTimingToFrom1.size(),interviewTimingToFrom2.size());
int maxSize = Math.max(interviewTimingToFrom1.size(),interviewTimingToFrom2.size());
StringBuilder result = new StringBuilder();
for (int i=0; i<maxSize; i++)
{
if (i < interviewTimingToFrom1.size())
result.append(interviewTimingToFrom1.get(i));
if (i < interviewTimingToFrom2.size())
result.append(interviewTimingToFrom2.get(i));
}
System.out.println(result.toString());
StringBuilder结果=新建StringBuilder();
对于(inti=0;i试试这个
List<String> interviewTimingToFrom1 = Arrays.asList(interviewTime1.split(","));
List<String> interviewTimingToFrom2 = Arrays.asList(interviewTime2.split(","));
if (interviewTimingToFrom1.size() == interviewTimingToFrom2.size()) {
int noOfSlots = interviewTimingToFrom1.size();
for (int i = 0; i < noOfSlots; i++) {
System.out.println("from " + interviewTimingToFrom1.get(i)
+ " to " + interviewTimingToFrom1.get(i));
}
} else {
System.out.println("No match");
int noOfSlots = (interviewTimingToFrom1.size() > interviewTimingToFrom2
.size() ? interviewTimingToFrom2.size()
: interviewTimingToFrom1.size());
for (int i = 0; i < noOfSlots; i++) {
System.out.println("from " + interviewTimingToFrom1.get(i)
+ " to " + interviewTimingToFrom2.get(i));
}
}
List interviewTimingToFrom1=Arrays.asList(interviewTime1.split(“,”);
List interviewTimingToFrom2=Arrays.asList(interviewTime2.split(“,”);
如果(interviewTimingToFrom1.size()==interviewTimingToFrom2.size()){
int noOfSlots=从1.size()开始的面试时间;
对于(int i=0;iinterviewTimingToFrom2
.size()?面试时间从2.size()开始
:面试时间从1.size()开始;
对于(int i=0;i
您是否考虑过对(int i=0,…)使用良好的旧
?我无法使用替代值生成该字符串如何生成?抱歉,忘记了该行上的实际方法调用。您得到的是两个列表之间的最大值。请检查我编辑的答案。@Nik您不认为它应该是min
?是的,它工作了,但我想将其设置为类似于从凌晨12点到凌晨1点,从凌晨2点到凌晨3点的字符串是的,你能吗help@lucifer连接它们。换句话说,使用+
操作符,如System.out.println(“foo”+值+“bar”+其他值);
@lucifer我再次编辑了我的答案。它现在将结果保存在StringBuilder
中,并在循环完成后输出结果。
List<String> interviewTimingToFrom1 = Arrays.asList(interviewTime1.split(","));
List<String> interviewTimingToFrom2 = Arrays.asList(interviewTime2.split(","));
if (interviewTimingToFrom1.size() == interviewTimingToFrom2.size()) {
int noOfSlots = interviewTimingToFrom1.size();
for (int i = 0; i < noOfSlots; i++) {
System.out.println("from " + interviewTimingToFrom1.get(i)
+ " to " + interviewTimingToFrom1.get(i));
}
} else {
System.out.println("No match");
int noOfSlots = (interviewTimingToFrom1.size() > interviewTimingToFrom2
.size() ? interviewTimingToFrom2.size()
: interviewTimingToFrom1.size());
for (int i = 0; i < noOfSlots; i++) {
System.out.println("from " + interviewTimingToFrom1.get(i)
+ " to " + interviewTimingToFrom2.get(i));
}
}