LeetCode Count Binary Substrings

Posted 2020-10-11 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/count-binary-substrings/description/

题目：

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.

题解：

把相邻的0或者1数目累计起来. 0001111, 就是3个0, 4个1. 当遇到下阶段或者string尾部. 取之前两组不同长度的最小值加进res中. res加上3 和 4的较小值.

Time Complexity: O(s.length()). Space: O(1).

AC Java:

 1 class Solution {
 2     public int countBinarySubstrings(String s) {
 3         if(s == null || s.length() == 0){
 4             return 0;
 5         }
 6         
 7         int res = 0;
 8         int preLen = 0;
 9         int curLen = 1;
10         for(int i = 1; i<s.length(); i++){
11             if(s.charAt(i) == s.charAt(i-1)){
12                 curLen++;
13             }else{
14                 res += Math.min(preLen, curLen);
15                 preLen = curLen;
16                 curLen = 1;
17             }
18         }
19         
20         res += Math.min(preLen, curLen);
21         return res;
22     }
23 }