leetcode: Add 0647

src/leetcode/0322.coin-change/main.py

@@ -2,14 +2,19 @@
 
 def coinChange(coins: List[int], amount: int) -> int:
     def dp(n):
-        # 基本情况
-        INIT_COUNT = 2 ** 10
+        # 基本常量
+        INITIAL_COUNT = 2 ** 30
         INVALID_COUNT = -1
+        # 初始条件
         if n == 0:
+            # 恰好兑换完
             return 0
         if n < 0:
+            # 说明没有合适的币值去完成兑换
             return INVALID_COUNT
-        min_exchange_count = INIT_COUNT
+
+        # 初始的最小兑换次数要足够大
+        min_exchange_count = INITIAL_COUNT
 
         # 依次遍历每一种币值, 并使用其中的币值换一次
         # 找出最小的兑换次数
@@ -20,7 +25,7 @@ def dp(n):
                 min_exchange_count = min(min_exchange_count, exchange_count)
 
         # 最后返回最小值
-        if min_exchange_count == INIT_COUNT:
+        if min_exchange_count == INITIAL_COUNT:
             return INVALID_COUNT
         else:
             return min_exchange_count

src/leetcode/0647.palindromic-substrings/Cargo.toml

@@ -0,0 +1,7 @@
+[package]
+name = "lc-0647-palindromic-substrings"
+version = "0.1.0"
+edition = "2021"
+publish = false
+
+[dependencies]

src/leetcode/0647.palindromic-substrings/index.md

@@ -0,0 +1,4 @@
+
+# 
+
+[问题描述](https://leetcode.com/problems/)

src/leetcode/0647.palindromic-substrings/main.py

@@ -0,0 +1,45 @@
+#!/usr/bin/env python3
+
+def countSubstrings(s: str) -> int:
+    # 回文字符串的一个特性是中间对齐, 我们可以遍历所有字符, 以它为中间, 尝试
+    # 向左右两侧扩展, 检查它们是否相等, 如果相等, 就说明这个 substring 是回文的.
+
+    palindrome_count = 0 
+
+    def twoSideEquals(s: str, left: int, right: int) -> bool:
+        # 先检查两个索引值是否有效, 再检查它们指向的字符是否相等
+        if left >= 0 and right < len(s) and s[left] == s[right]:
+            return True
+        else:
+            return False
+
+    # middle 是 substring 的中间节点
+    for middle in range(len(s)):
+        # 如果 substring 中包含奇数个字符, 那就以 (middle) 作为中间对齐节点
+        left = middle
+        right = middle
+        while twoSideEquals(s, left, right):
+            palindrome_count += 1
+            left -= 1
+            right += 1
+
+        # 如果 substring 中包含偶数个字符, 那就以 (middle, middle+1) 作为中间对齐节点
+        left = middle
+        right = middle + 1
+        while twoSideEquals(s, left, right):
+            palindrome_count += 1
+            left -= 1
+            right += 1
+
+    # 返回结果
+    return palindrome_count
+
+def main():
+    s1 = "abc"
+    assert(countSubstrings(s1) == 3)
+
+    s2 = "aaa"
+    assert(countSubstrings(s2) == 6)
+
+if __name__ == "__main__":
+    main()

src/leetcode/0647.palindromic-substrings/src/main.rs

@@ -0,0 +1,27 @@
+// Copyright (c) 2024 Xu Shaohua <[email protected]>. All rights reserved.
+// Use of this source is governed by General Public License that can be found
+// in the LICENSE file.
+
+pub fn solution1() {
+    todo!();
+}
+
+pub type SolutionFn = fn();
+
+fn check_solution(_func: SolutionFn) {
+    todo!();
+}
+
+fn main() {
+    check_solution(solution1);
+}
+
+#[cfg(test)]
+mod tests {
+    use super::{check_solution, solution1};
+
+    #[test]
+    fn test_solution1() {
+        check_solution(solution1);
+    }
+}