LeetCode——Count and Say
2014-11-12 18:25
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The count-and-say sequence is the sequence of integers beginning as follows:
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string。
原题链接:https://oj.leetcode.com/problems/count-and-say/
可以看出,后一个数字是前个数字的读法,21读作1个2,先写下12,1个1,再写下11,连起来1211.
public static String countAndSay(int n) {
if(n <= 0)
return null;
String str = "1";
int count = 1;
for(int i=0;i<n-1;i++){
StringBuilder builder = new StringBuilder();
for(int j=0;j<str.length();j++){
if(j<str.length()-1 && str.charAt(j)==str.charAt(j+1))
count++;
else{
builder.append(count + "" + str.charAt(j));
count = 1;
}
}
str = builder.toString();
}
return str;
}
1, 11, 21, 1211, 111221, ...
1is read off as
"one 1"or
11.
11is read off as
"two 1s"or
21.
21is read off as
"one 2, then
one 1"or
1211.
Given an integer n, generate the nth sequence.
Note: The sequence of integers will be represented as a string。
原题链接:https://oj.leetcode.com/problems/count-and-say/
可以看出,后一个数字是前个数字的读法,21读作1个2,先写下12,1个1,再写下11,连起来1211.
public static String countAndSay(int n) {
if(n <= 0)
return null;
String str = "1";
int count = 1;
for(int i=0;i<n-1;i++){
StringBuilder builder = new StringBuilder();
for(int j=0;j<str.length();j++){
if(j<str.length()-1 && str.charAt(j)==str.charAt(j+1))
count++;
else{
builder.append(count + "" + str.charAt(j));
count = 1;
}
}
str = builder.toString();
}
return str;
}
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