696. 计数二进制子串 Krains 2020-08-10 09:42:23 字符串

class Solution {
    public int countBinarySubstrings(String s) {
        int ans = 0;
        int i = 0;
        // pre表示之前一段连续1或0个数，cur为当前连续0或1的个数
        int pre = 0, cur = 0;
        while(i < s.length()){
            char c = s.charAt(i);
            while(i < s.length() && s.charAt(i) == c){
                i++;
                cur++;
            }
            ans += Math.min(pre, cur);
            pre = cur;
            cur = 0;
        }
        
        return ans;
    }
}

class Solution {
    public int countBinarySubstrings(String s) {
        int ans = 0;
        int pre = 0, cur = 1;
        for(int i = 1; i < s.length(); i++){
            if(s.charAt(i-1) == s.charAt(i)){
                cur++;
            }else{
                ans += Math.min(pre, cur);
                pre = cur;
                cur = 1;
            }
        }
        // 加上末尾未统计的
        ans += Math.min(pre, cur);

        return ans;
    }
}