一刷434

Author: diguage

README.adoc

@@ -3064,13 +3064,13 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 |Medium
 |
 
-//|{counter:codes}
-//|{leetcode_base_url}/number-of-segments-in-a-string/[434. Number of Segments in a String^]
-//|{source_base_url}/_0434_NumberOfSegmentsInAString.java[Java]
-//|{doc_base_url}/0434-number-of-segments-in-a-string.adoc[题解]
-//|Easy
-//|
-//
+|{counter:codes}
+|{leetcode_base_url}/number-of-segments-in-a-string/[434. Number of Segments in a String^]
+|{source_base_url}/_0434_NumberOfSegmentsInAString.java[Java]
+|{doc_base_url}/0434-number-of-segments-in-a-string.adoc[题解]
+|Easy
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/non-overlapping-intervals/[435. Non-overlapping Intervals^]
 //|{source_base_url}/_0435_NonOverlappingIntervals.java[Java]

docs/0434-number-of-segments-in-a-string.adoc

@@ -1,24 +1,49 @@
 [#0434-number-of-segments-in-a-string]
-= 434. Number of Segments in a String
+= 434. 字符串中的单词数
 
-{leetcode}/problems/number-of-segments-in-a-string/[LeetCode - Number of Segments in a String^]
+https://leetcode.cn/problems/number-of-segments-in-a-string/[LeetCode - 434. 字符串中的单词数 ^]
 
-Count the number of segments in a string, where a segment is defined to be a contiguous sequence of non-space characters.
+统计字符串中的单词个数，这里的单词指的是连续的不是空格的字符。
 
-Please note that the string does not contain any *non-printable* characters.
+请注意，你可以假定字符串里不包括任何不可打印的字符。
+
+*示例:*
+
+....
+输入: "Hello, my name is John"
+输出: 5
+解释: 这里的单词是指连续的不是空格的字符，所以 "Hello," 算作 1 个单词。
+....
 
-*Example:*
-[subs="verbatim,quotes,macros"]
-----
-*Input:* "Hello, my name is John"
-*Output:* 5
-----
 
+== 思路分析
 
+识别出单词开始的字符，进行统计数量，其余向后走。
 
 [[src-0434]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0434_NumberOfSegmentsInAString.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0434_NumberOfSegmentsInAString_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
 
+. https://leetcode.cn/problems/number-of-segments-in-a-string/solutions/1034164/gong-shui-san-xie-jian-dan-zi-fu-mo-ni-t-0gx6/[434. 字符串中的单词数 - 简单字符模拟题^]
+. https://leetcode.cn/problems/number-of-segments-in-a-string/solutions/1033996/zi-fu-chuan-zhong-de-dan-ci-shu-by-leetc-igfb/[434. 字符串中的单词数 - 官方题解^]

docs/index.adoc

@@ -953,7 +953,7 @@ include::0429-n-ary-tree-level-order-traversal.adoc[leveloffset=+1]
 
 include::0433-minimum-genetic-mutation.adoc[leveloffset=+1]
 
-// include::0434-number-of-segments-in-a-string.adoc[leveloffset=+1]
+include::0434-number-of-segments-in-a-string.adoc[leveloffset=+1]
 
 // include::0435-non-overlapping-intervals.adoc[leveloffset=+1]
 

logbook/202503.adoc

@@ -1314,6 +1314,11 @@ endif::[]
 |{doc_base_url}/0423-reconstruct-original-digits-from-english.adoc[题解]
 |⭕️ 按照最初想法，先统计每个字符出现的次数，然后再按照数字顺序，逐个去尝试是否在字符串中。这样时间复杂度就不可控。看题解，可以按照单词中的字符出现特性，先将只出现在一个数字中的字符挑选出来，确定对应的数字出现次数。然后，在处理出现中两个数字中的字符，逐步把所有数字出现次数确定好。
 
+|{counter:codes2503}
+|{leetcode_base_url}/number-of-segments-in-a-string/[434. 字符串中的单词数^]
+|{doc_base_url}/0434-number-of-segments-in-a-string.adoc[题解]
+|✅ 识别单词开头，进行数量统计。
+
 
 |===
 

src/main/java/com/diguage/algo/leetcode/_0434_NumberOfSegmentsInAString.java

@@ -0,0 +1,25 @@
+package com.diguage.algo.leetcode;
+
+public class _0434_NumberOfSegmentsInAString {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-08-29 20:49:07
+   */
+  public int countSegments(String s) {
+    int result = 0;
+    boolean flag = false;
+    for (int i = 0; i < s.length(); i++) {
+      if (s.charAt(i) == ' ') {
+        flag = false;
+        continue;
+      }
+      if (!flag) {
+        result++;
+        flag = true;
+      }
+    }
+    return result;
+  }
+  // end::answer[]
+}