696. Count Binary Substrings

这道题，最重要的是要能观察出，连续的0和连续的1之间的关系——每一组连续的0和连续的1可以贡献出：Math.min(连续0，连续1)

下面的两个算法都可以beat 100%，时间复杂度O(n).

    public int countBinarySubstrings(String s) {
        int res = 0;
        int cur = 1, pre = 0;
        char[] cs = s.toCharArray();
        for (int i = 1; i < cs.length; i++) {
            if (cs[i] == cs[i - 1])
                cur++;
            else {
                res += Math.min(pre, cur);
                pre = cur;
                cur = 1;
            }
        }
        return res + Math.min(pre, cur);
    }

    public int countBinarySubstrings(String s) {
        int res = 0;
        int cur = 1, pre = 0;
        char[] cs = s.toCharArray();
        for (int i = 1; i < cs.length; i++) {
            if (cs[i] == cs[i - 1])
                cur++;
            else {
                pre = cur;
                cur = 1;
            }
            if (pre >= cur)
                res++;
        }
        return res;
    }

标签：Count,Binary,pre,cur,int,res,++,Substrings,cs
来源： https://www.cnblogs.com/feiflytech/p/15824900.html