696. Count Binary Substrings

Question

696. Count Binary Substrings

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.

Solution

思路:题目大意是,给一个二进制的字符串，问有多少子串的0个数量等于1的数量且子串中0和1不能交替出现。

00110011
  zeroCount oneCount last
0 1         0         -1->0  
0 2         0         0
1 2         1         0->1      有一个符合要求的子串01
1 2         2         1         有一个符合要求的子串0011
0 1         2         1->0      有一个符合要求的子串10
0 2         2         0         有一个符合要求的子串1100
1 2         1         0->1      有一个符合要求的子串01
1 2         2         1         有一个符合要求的子串0011

Java实现:

public int countBinarySubstrings(String s) {
    // 子串满足
    // 1)0和1个数相等
    // 2)0与1不能交替出现

    int ans = 0;
    int zeroCount = 0;
    int oneCount = 0;
    int last = -1;
    for (int i=0; i<s.length(); i++) {
        char c = s.charAt(i);
        if (c == '0') {
            if (last == -1 || last == 1) {
                zeroCount = 0;
                last = 0;
            }
            zeroCount++;
        } else {
            if (last == -1 || last == 0) {
                oneCount = 0;
                last = 1;
            }
            oneCount++;
        }
        if (last == 0) ans += zeroCount<=oneCount?1:0;
        else ans += oneCount<=zeroCount?1:0;
    }
    return ans;
}