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《LeetCode刷题(Easy Rank):696. Count Binary Substrings》

 3月前  •   LeetCode  •     •   19  •   0

Question:

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:

示例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:

示例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.

Solution:

字符串“000111”有3个子串,字符串“0001111”也是3个子串,规律:组成的子串数目由连续0的数目和连续1的数目中较小的决定。所以此题的解题关键就是统计连续0的数目和连续1的数目。

JAVAclass Solution {
    public int countBinarySubstrings(String s) {
        int len = s.length();
        if(len<2){
            return 0;
        }
        char[] chs = s.toCharArray();
        int count=1;
        List<Integer> list = new ArrayList<>();
        for(int i=1;i<len;i++){
            if(chs[i]!=chs[i-1]){
                list.add(count);//统计连续0的数目和连续1的数目,将结果放入list中
                count=0;
            }
            count++;
        }
        int result = 0;
        list.add(count);
        for(int i=0;i<list.size()-1;i++){
            result += Math.min(list.get(i),list.get(i+1));//结果由较小的那部分决定
        }
        return result;
    }
}

 

 

 

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