计算字符串中重复的元素

Question

题目是求一个字符'*'在一个字符串中连续出现的次数。我已经为它写了一个程序，但是有没有其他有效的方法来解决这个问题

        for(int i=0;i<n;i++){
            if(s.charAt(i) == '*'){
                count=1;
                for(int j=i+1;j<n;j++){
                    if(s.charAt(j) != s.charAt(i))
                        break;
                    else
                        count++;
                }
            }
        }

例如：

input: *a**b  output should be 2
input: ***a**b output should be 3

Answer 1

简单一点，线性复杂度（此任务的最高效率）：

int stars = 0; 
int longest = 0;
for(int i=0;i<n;i++){
        if(s.charAt(i) == '*') {
            stars++;
            longest = Math.max(longest, stars);
            //here you can do something useful - 
            // for example, get max number of consecutive stars
        }
        else {
            //here you can do something useful - in case of  if (stars > 0) 
            // for example - increment number of series
            //and don't forget to do the same action after loop finish
            stars = 0;
        }      
}

计算字符串中重复的元素

count repeated elements in a string

java

string

algorithm

performance

time-complexity