From e035548e40b9e056e18c87281e2b1ee1279213c1 Mon Sep 17 00:00:00 2001
From: CyC2018 <zhengyc101@163.com>
Date: Wed, 28 Oct 2026 13:45:07 +0800
Subject: [PATCH] =?UTF-8?q?Update=2057.2=20=E5=92=8C=E4=B8=BA=20S=20?=
 =?UTF-8?q?=E7=9A=84=E8=BF=9E=E7=BB=AD=E6=AD=A3=E6=95=B0=E5=BA=8F=E5=88=97?=
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题目链接改为力扣中的对应题目，题目描述和代码按该题的接口调整。
---
 notes/57.2 和为 S 的连续正数序列.md | 25 +++++++++++++------------
 1 file changed, 13 insertions(+), 12 deletions(-)

diff --git a/notes/57.2 和为 S 的连续正数序列.md b/notes/57.2 和为 S 的连续正数序列.md
index f09420f2..5ab87287 100644
--- a/notes/57.2 和为 S 的连续正数序列.md	
+++ b/notes/57.2 和为 S 的连续正数序列.md	
@@ -1,8 +1,8 @@
 # 57.2 和为 S 的连续正数序列
 
-## 题目描述
+## 题目链接
 
-[牛客网](https://www.nowcoder.com/practice/c451a3fd84b64cb19485dad758a55ebe?tpId=13&tqId=11194&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/he-wei-sde-lian-xu-zheng-shu-xu-lie-lcof/)
 
 ## 题目描述
 
@@ -16,28 +16,29 @@
 ## 解题思路
 
 ```java
-public ArrayList<ArrayList<Integer>> FindContinuousSequence(int sum) {
-    ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
+public int[][] fileCombination(int target) {
+    List<int[]> ret = new ArrayList<>();
     int start = 1, end = 2;
     int curSum = 3;
-    while (end < sum) {
-        if (curSum > sum) {
+    while (end < target) {
+        if (curSum > target) {
             curSum -= start;
             start++;
-        } else if (curSum < sum) {
+        } else if (curSum < target) {
             end++;
             curSum += end;
         } else {
-            ArrayList<Integer> list = new ArrayList<>();
-            for (int i = start; i <= end; i++)
-                list.add(i);
-            ret.add(list);
+            int[] arr = new int[end - start + 1];
+            for (int i = start; i <= end; i++) {
+                arr[i - start] = i;
+            }
+            ret.add(arr);
             curSum -= start;
             start++;
             end++;
             curSum += end;
         }
     }
-    return ret;
+    return ret.toArray(new int[ret.size()][]);
 }
 ```