标签：并且 题目 ++ bec 字符串 val 运行时间 notice 运行时
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.1 class Solution { 2 public int countBinarySubstrings(String s) { 3 char[] array = s.toCharArray(); 4 int ans = 0; 5 int pre = 0; //记录前一个字符出现的次数 6 int cur = 1; //记录后一个字符出现的次数 7 for ( int i = 1 ; i < array.length ; i ++ ){ 8 if ( array[i] == array[i-1] ) cur++; 9 else { 10 pre = cur; 11 cur = 1; 12 } 13 if ( pre >= cur ) ans++; 14 } 15 return ans; 16 } 17 }
运行时间12ms，击败99.85%的人。
[leetcode] Count Binary Substrings
标签：并且 题目 ++ bec 字符串 val 运行时间 notice 运行时
原文地址：https://www.cnblogs.com/boris1221/p/9304122.html
