zoukankan      html  css  js  c++  java
  • 696. 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.

    Approach #1: String. [Java]

    class Solution {
        public int countBinarySubstrings(String s) {
            int l = s.length();
            int ret = 0;
            int preRunLength = 0;
            int curRunLength = 1;
            char[] ch = s.toCharArray();
            for (int i = 1; i < l; ++i) {
                if (ch[i] == ch[i-1]) curRunLength++;
                else {
                    preRunLength = curRunLength;
                    curRunLength = 1;
                }
                if (preRunLength >= curRunLength) ret++;
            }
            return ret;
        }
    }
    

      

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    【bzoj1191】 HNOI2006—超级英雄Hero
    【poj3020】 Antenna Placement
    【poj1274】 The Perfect Stall
    【poj2724】 Purifying Machine
    【poj2226】 Muddy Fields
    【codevs1257】 打砖块
    【poj2186】 Popular Cows
    【poj1236】 Network of Schools
    【poj1144】 Network
    【poj3177】 Redundant Paths
  • 原文地址:https://www.cnblogs.com/h-hkai/p/10771438.html
Copyright © 2011-2022 走看看