再次完成560

README.adoc

@@ -3928,7 +3928,7 @@
 |{source_base_url}/_0560_SubarraySumEqualsK.java[Java]
 |{doc_base_url}/0560-subarray-sum-equals-k.adoc[Note]
 |Medium
-|
+|前缀和
 
 //|561
 //|{leetcode_base_url}/array-partition-i/[Array Partition I]

docs/0560-subarray-sum-equals-k.adoc

@@ -25,17 +25,40 @@ Given an array of integers and an integer **k**, you need to find the total numb
 
 在完成遍历数组后，`count` 记录了所需结果
 
+基于一个idea：`sum[j] - sum[i] == k` 的话，`nums[i+1, j]` 之间数字的和就是 `k`。
+
+image::images/0560-01.png[]
+
+image::images/0560-02.png[]
+
+image::images/0560-03.png[]
+
+image::images/0560-04.png[]
+
+image::images/0560-05.png[]
+
+image::images/0560-06.png[]
+
+image::images/0560-07.png[]
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+image::images/0560-08.png[]
+
+image::images/0560-09.png[]
+
 [[src-0560]]
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0560_SubarraySumEqualsK.java[]
 ----
 
-这个方法是基于一个idea：`sum[j] - sum[i] == k` 的话，`nums[i, j ]` 之间数字的和就是 `k`。
+[{java_src_attr}]
+----
+include::{sourcedir}/_0560_SubarraySumEqualsK_2.java[]
+----
 
 
 == 参考资料
 
-. https://leetcode-cn.com/problems/subarray-sum-equals-k/solution/he-wei-kde-zi-shu-zu-by-leetcode/[和为K的子数组 - 和为K的子数组 - 力扣（LeetCode）]
+. https://leetcode.cn/problems/continuous-subarray-sum/solutions/807930/lian-xu-de-zi-shu-zu-he-by-leetcode-solu-rdzi/[和为K的子数组 - 和为K的子数组 - 官方题解^]
 
 

src/main/java/com/diguage/algorithm/leetcode/_0560_SubarraySumEqualsK.java

@@ -9,7 +9,7 @@
  *
  * https://leetcode.com/problems/subarray-sum-equals-k/[Subarray Sum Equals K - LeetCode]
  *
- * @author D瓜哥, https://www.diguage.com/
+ * @author D瓜哥 · https://www.diguage.com
  * @since 2020-01-30 23:14
  */
 public class _0560_SubarraySumEqualsK {

src/main/java/com/diguage/algorithm/leetcode/_0560_SubarraySumEqualsK_2.java

@@ -0,0 +1,59 @@
+package com.diguage.algorithm.leetcode;
+
+import java.util.*;
+
+/**
+ * = 560. Subarray Sum Equals K
+ *
+ * https://leetcode.com/problems/subarray-sum-equals-k/[Subarray Sum Equals K - LeetCode]
+ *
+ * @author D瓜哥 · https://www.diguage.com
+ * @since 2024-06-23 11:08:29
+ */
+public class _0560_SubarraySumEqualsK_2 {
+    /**
+     * 自己根据“前缀和”套路想的思路，参考 https://leetcode.cn/problems/subarray-sum-equals-k/solutions/238572/he-wei-kde-zi-shu-zu-by-leetcode-solution/[560. 和为 K 的子数组 - 官方题解^] 更正了代码。
+     */
+    public int subarraySum(int[] nums, int k) {
+      int result = 0;
+      if (nums == null || nums.length == 0) {
+        return result;
+      }
+      // key：前缀和，value：key 对应的前缀和的个数
+      Map<Integer, Integer> sumToCntMap = new HashMap<>();
+      // 对于下标为 0 的元素，前缀和为 0，个数为 1
+      sumToCntMap.put(0, 1);
+      int sum = 0;
+      for (int i = 0; i < nums.length; i++) {
+        sum += nums[i];
+        // 先获得前缀和为 preSum - k 的个数，加到计数变量里
+        // TODO 这里为什么先检查是否存在？
+        if (sumToCntMap.containsKey(sum - k)) {
+          result += sumToCntMap.get(sum - k);
+        }
+        // 然后维护 preSumFreq 的定义
+        sumToCntMap.put(sum, sumToCntMap.getOrDefault(sum, 0) + 1);
+      }
+      return result;
+    }
+
+    public static void main(String[] args) {
+        _0560_SubarraySumEqualsK_2 solution = new _0560_SubarraySumEqualsK_2();
+
+        int[] n4 = {3, 4, 7, 2, -3, 1, 4, 2};
+        int r4 = solution.subarraySum(n4, 7);
+        System.out.println((r4 == 4) + " : " + r4);
+
+        int[] n3 = {1};
+        int r3 = solution.subarraySum(n3, 0);
+        System.out.println((r3 == 0) + " : " + r3);
+
+        int[] n2 = {-1, -1, 1};
+        int r2 = solution.subarraySum(n2, 0);
+        System.out.println((r2 == 1) + " : " + r2);
+
+        int[] n1 = {1, 1, 1};
+        int r1 = solution.subarraySum(n1, 2);
+        System.out.println((r1 == 2) + " : " + r1);
+    }
+}