Topic

• String

Description

https://leetcode.com/problems/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.

Analysis

方法一：暴力算法，遇到超長字符串會超時，故棄。

方法二：發現字符串其中規律

如’00000111’，一個連續’0’與連續’1’的字符串，符合題意要求的子字符串為min('0‘的個數, '1’的個數)=min(5,3)=3。

如’000011100111‘字符串，以連續相同字符分為一個區間規則，得到[0,3]，[4,6]，[7,8]，[9,11]共4個區間。

由2. 得到區間，將其兩兩相鄰配對，如[0,3]與[4,6]，[4,6]與[7,8]，再由1. 得出規律求出兩配對區間符合題意要求的子字符串個數，最后將其累加，便能得到最后結果。

方法三：方法二改進版。

Submission

import java.util.ArrayList; import java.util.List;public class CountBinarySubstrings {// 方法一：暴力算法，遇到超長字符串會超時public int countBinarySubstrings(String s) {int result = 0;// List<String> resultList = new ArrayList<>();for (int i = 0; i < s.length(); i++) {int state = 1;char tempChar = s.charAt(i);int firstCharCount = 1, secondCharCount = 0, rightPartCount = 0;for (int j = i + 1; j < s.length(); j++) {if (state == 1) {if (tempChar == s.charAt(j)) {firstCharCount++;} else {state = 2;tempChar = s.charAt(j);}}if (state == 2) {if (tempChar == s.charAt(j)) {secondCharCount++;}rightPartCount++;if (rightPartCount == firstCharCount) {if (secondCharCount == firstCharCount) {result++;// resultList.add(repeat(tempChar=='1'?'0':'1', firstCharCount) +// repeat(tempChar, secondCharCount));}break;}}}}return result;}// 方法二：public int countBinarySubstrings2(String s) {List<int[]> list = new ArrayList<>();int result = 0, tempIndex = 0;for (int i = tempIndex + 1; i <= s.length(); i++) {if (i == s.length() || s.charAt(i) != s.charAt(tempIndex)) {list.add(new int[] { tempIndex, i - 1 });tempIndex = i;}}for (int i = 0; i < list.size(); i++) {int[] leftPart = list.get(i);if (i + 1 == list.size()) {break;}int[] rightPart = list.get(i + 1);int leftSize = leftPart[1] - leftPart[0] + 1;int rightSize = rightPart[1] - rightPart[0] + 1;result += Math.min(leftSize, rightSize);}return result;}// 方法三：方法二的改進版public int countBinarySubstrings3(String s) {int result = 0, lastIndex = 0, lastSize = 0;for (int i = lastIndex + 1; i <= s.length(); i++) {if (i == s.length() || s.charAt(i) != s.charAt(lastIndex)) {if (lastSize == 0) {lastSize = i - lastIndex;} else {int currentSize = i - lastIndex;result += Math.min(lastSize, currentSize);lastSize = currentSize;}lastIndex = i;}}return result;}private String repeat(char c, int times) {StringBuilder sb = new StringBuilder();for (int i = 0; i < times; i++) {sb.append(c);}return sb.toString();}}

Test

import static org.junit.Assert.*; import org.junit.Test;public class CountBinarySubstringsTest {@Testpublic void test() {CountBinarySubstrings obj = new CountBinarySubstrings();assertEquals(6, obj.countBinarySubstrings("00110011"));assertEquals(4, obj.countBinarySubstrings("10101"));}@Testpublic void test2() {CountBinarySubstrings obj = new CountBinarySubstrings();assertEquals(6, obj.countBinarySubstrings2("00110011"));assertEquals(4, obj.countBinarySubstrings2("10101"));}@Testpublic void test3() {CountBinarySubstrings obj = new CountBinarySubstrings();assertEquals(0, obj.countBinarySubstrings3("00"));assertEquals(1, obj.countBinarySubstrings3("001"));assertEquals(6, obj.countBinarySubstrings3("00110011"));assertEquals(4, obj.countBinarySubstrings3("10101"));assertEquals(3, obj.countBinarySubstrings3("00110"));} }

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