完成523

Author: diguage

docs/0000-00-note.adoc

@@ -2,3 +2,9 @@
 
 . xref:0215-kth-largest-element-in-an-array.adoc[215. Kth Largest Element in an Array] 没想到快速排序的分区算法，竟然可以用于做快速选择？！神奇… 可惜的是，这题目以前做过，现在都给忘完了…
 . xref:0437-path-sum-iii.adoc[437. Path Sum III] 前缀和的解法还需要再思考思考！
+
+== 解题技巧
+
+. 树形 DP 套路
+. 前缀和
+.. xref:0523-continuous-subarray-sum.adoc[523. Continuous Subarray Sum]

docs/0523-continuous-subarray-sum.adoc

@@ -1,3 +1,4 @@
+[#0523-continuous-subarray-sum]
 = 523. Continuous Subarray Sum
 
 https://leetcode.com/problems/continuous-subarray-sum/[LeetCode - Continuous Subarray Sum]
@@ -33,6 +34,24 @@ Given a list of *non-negative* numbers and a target *integer* k, write a functio
 . You may assume the sum of all the numbers is in the range of a signed 32-bit integer.
 
 
+== 解题分析
+
+利用同余定理：
+
+image::images/0523-01.png[]
+
+当 latexmath:[prefixSums[q\]−prefixSums[p\]] 为 latexmath:[k] 的倍数时，latexmath:[prefixSums[p\]] 和 latexmath:[prefixSums[q\]] 除以 latexmath:[k] 的余数相同。（_D瓜哥注：余数相同，则相减之后余数就被减掉了。_）因此只需要计算每个下标对应的前缀和除以 latexmath:[k] 的余数即可，使用哈希表存储每个余数第一次出现的下标。
+
+image::images/0523-02.png[]
+
+image::images/0523-03.png[]
+
+image::images/0523-04.png[]
+
+image::images/0523-05.png[]
+
+image::images/0523-06.png[]
+
 
 
 [[src-0523]]

docs/images/0523-01.png

@@ -10,7 +10,7 @@
  *
  * https://leetcode.com/problems/balanced-binary-tree/[LeetCode - Balanced Binary Tree]
  *
- * @author D瓜哥, https://www.diguage.com/
+ * @author D瓜哥 · https://www.diguage.com
  * @since 2020-02-06 23:10
  */
 public class _0110_BalancedBinaryTree_21 {

docs/images/0523-02.png

@@ -0,0 +1,45 @@
+package com.diguage.algorithm.leetcode;
+
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+ * @author D瓜哥 · https://www.diguage.com
+ * @since 2024-06-23 00:23:40
+ */
+
+public class _0523_ContinuousSubarraySum {
+  /**
+   * 参考 https://leetcode.cn/problems/continuous-subarray-sum/solutions/807930/lian-xu-de-zi-shu-zu-he-by-leetcode-solu-rdzi/[523. 连续的子数组和 - 官方题解^]
+   */
+  public boolean checkSubarraySum(int[] nums, int k) {
+    if (nums == null || nums.length < 2) {
+      return false;
+    }
+    Map<Integer, Integer> remainderToIndexMap = new HashMap<>();
+    // 当 sum=0 时，还没有任何数字参与，所以是 -1
+    remainderToIndexMap.put(0, -1);
+    int remainder = 0;
+    for (int i = 0; i < nums.length; i++) {
+      // 同余定理
+      remainder = (remainder + nums[i]) % k;
+      if (remainderToIndexMap.containsKey(remainder)) {
+        int prevIndex = remainderToIndexMap.get(remainder);
+        if (i - prevIndex >= 2) {
+          return true;
+        }
+      } else {
+        // 前面已经判断过是否包含该值，
+        // 这里也就不需要为了保存最小的下标去判断是否已经存在该值
+        remainderToIndexMap.put(remainder, i);
+      }
+    }
+    return false;
+  }
+
+  public static void main(String[] args) {
+    _0523_ContinuousSubarraySum solution = new _0523_ContinuousSubarraySum();
+    boolean result = solution.checkSubarraySum(new int[]{5, 0, 0, 0}, 3);
+    System.out.println(result);
+  }
+}