[LintCode/LeetCode] Count Binary Substrings

BaronZhang 发布于2019-08-16 12:44 / 1517人阅读

Problem

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

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

public class Solution {
    /**
     * @param s: a string
     * @return: the number of substrings
     */
    public int countBinarySubstrings(String s) {
        // Write your code here
        int countZero = 0, countOne = 0, result = 0;
        char[] str = s.toCharArray();
        for (int i = 0; i < str.length; i++) {
            if (i == 0) {
                if (str[i] == "0") {
                    countZero++;
                } else countOne++;
            } else {
                if (str[i] == "0") {
                    if (str[i-1] == "0") countZero++;
                    else countZero = 1;
                    if (countOne >= countZero) result++;
                } else {
                    if (str[i-1] == "1") countOne++;
                    else countOne = 1;
                    if (countZero >= countOne) result++;
                }
            }
        }
        return result;
    }
}

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