test

src/leecode/java/3333.找到初始输入字符串 II.java

@@ -0,0 +1,81 @@
+/*
+ * @lc app=leetcode.cn id=3333 lang=java
+ * @lcpr version=30201
+ *
+ * [3333] 找到初始输入字符串 II
+ */
+
+// @lc code=start
+
+import java.util.ArrayList;
+import java.util.List;
+
+class Solution {
+    public int possibleStringCount(String word, int k) {
+        // 将原字符串转为数组，单字母连续出现的次数为数组中的元素，遇到一个新字母结算一个数组元素[]{a:3,b:3,a:4} {3,3,4}
+        // k的最小值即为数组长度，记为m，即至少每个出现连续块的字母取一次
+        // k向上增加时，即k-m个数需要被分配，此为递归
+        List<Integer> wordArr = new ArrayList<>();
+        int i = 0;
+        int sum = 0;
+        char c = word.charAt(i);
+        while (i < word.length()) {
+            if (word.charAt(i) != c) {
+                wordArr.add(sum);
+                c = word.charAt(i);
+                sum = 0;
+            }
+            sum++;
+            i++;
+        }
+        // 最多再分配k - wordArr.size()
+        int result = 0;
+        for (int remain = 0; remain < word.length(); remain++) {
+            result += dfs(wordArr, 0, word.length() - wordArr.size());
+        }
+        return result % (10^9+7);
+    }
+
+    private int dfs(List<Integer> wordArr, int index, int remain) {
+        // 字母分配完毕
+        if (remain == 0) {
+            return 1;
+        }
+        // 不可再分配
+        if (index >= wordArr.size()) {
+            return 0;
+        }
+        int sum = 0;
+        for (int i = 0; i <= wordArr.get(index); i++) {
+            sum += dfs(wordArr, index + 1, remain - i);
+        }
+        return sum;
+    }
+
+    public static void main(String[] args) throws InterruptedException {
+//        No3333 no = new No3333();
+//        System.out.println(no.possibleStringCount("aaabbb", 3));
+        while (true) {
+            Thread.sleep(1000);
+        }
+    }
+}
+// @lc code=end
+
+
+
+/*
+// @lcpr case=start
+// "aabbccdd"\n7\n
+// @lcpr case=end
+
+// @lcpr case=start
+// "aabbccdd"\n8\n
+// @lcpr case=end
+
+// @lcpr case=start
+// "aaabbb"\n3\n
+// @lcpr case=end
+
+ */
+