[leetcode] Count Binary Substrings

Give a string s, count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively.

Substrings that occur multiple times are counted the number of times they occur.

Example 1:

Input: "00110011"
Output: 6
Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".

Notice that some of these substrings repeat and are counted the number of times they occur.

Also, "00110011" is not a valid substring because all the 0's (and 1's) are not grouped together.

Example 2:

Input: "10101"
Output: 4
Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.

Note:

• s.length will be between 1 and 50,000.
• s will only consist of "0" or "1" characters.

---

 分析：题目的意思很好理解，字符串s由0、1组成，要求找满足条件的字串的个数，条件是：字串中包含相同数量的0和1，并且所有的0和1都要紧挨着。

      之前好像做过一个类似的题目，但是那个题要求找间隔开的。思路还是有点类似的。在这个题目里，因为要0和1数量相同，很容易想到用两个变量保存，pre用于保存前一个字符出现的次数，cur用来保存后一个字符出现的次数。整理思路后，代码如下：

class Solution {
   public int countBinarySubstrings(String s) {
        char[] array = s.toCharArray();
        int ans = 0;
        int pre = 0; //记录前一个字符出现的次数
        int cur = 1; //记录后一个字符出现的次数
        for ( int i = 1 ; i < array.length ; i ++ ){
            if ( array[i] == array[i-1] ) cur++;
            else {
                pre = cur;
                cur = 1;
            }
            if ( pre >= cur ) ans++;
        }
        return ans;
    }
}

      运行时间12ms，击败99.85%的人。