更新了[lcof-57.和为s的连续正整数序列]的cpp解法 (doocs#347)

* 增加了[lcof-18.删除链表的节点]和[lcof-40.最小的k个数]的cpp解法,基本双优

* 优化[lcof-40.最小的k个数]格式

* 更新了[lcof-57.和为s的连续正整数序列]的cpp版本解法

Co-authored-by: junfeng.fj <junfeng.fj@alibaba-inc.com>

Author: ChunelFeng

lcof/面试题57 - II. 和为s的连续正数序列/README.md

@@ -113,6 +113,55 @@ var findContinuousSequence = function (target) {
 };
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> build(int small, int big) {
+        vector<int> ret;
+        for (int i = small; i <= big; i++) {
+            ret.push_back(i);
+        }
+
+        return ret;
+    }
+
+    vector<vector<int>> findContinuousSequence(int target) {
+        vector<vector<int>> ret;
+        int small = 1;
+        int big = 2;
+        int mid = (target + 1) / 2;
+        int curSum = small + big;
+
+        if (target < 3) {
+            ret;
+        }
+
+        while(small < mid) {
+            if (curSum == target) {
+                ret.push_back(build(small, big));
+            }
+
+            while (curSum > target && small < mid) {
+                // 一直减去，减去到比target小停止
+                curSum -= small;
+                small++;
+
+                if (curSum == target && small < mid) {
+                    ret.push_back(build(small, big));
+                }
+            }
+
+            big++;
+            curSum += big;
+        }
+
+        return ret;
+    }
+};
+```
+
 ### **...**
 
 ```

lcof/面试题57 - II. 和为s的连续正数序列/Solution.cpp

@@ -0,0 +1,44 @@
+class Solution {
+public:
+    vector<int> build(int small, int big) {
+        vector<int> ret;
+        for (int i = small; i <= big; i++) {
+            ret.push_back(i);
+        }
+
+        return ret;
+    }
+
+    vector<vector<int>> findContinuousSequence(int target) {
+        vector<vector<int>> ret;
+        int small = 1;
+        int big = 2;
+        int mid = (target + 1) / 2;
+        int curSum = small + big;
+
+        if (target < 3) {
+            ret;
+        }
+
+        while(small < mid) {
+            if (curSum == target) {
+                ret.push_back(build(small, big));
+            }
+
+            while (curSum > target && small < mid) {
+                // 一直减去，减去到比target小停止
+                curSum -= small;
+                small++;
+
+                if (curSum == target && small < mid) {
+                    ret.push_back(build(small, big));
+                }
+            }
+
+            big++;
+            curSum += big;
+        }
+
+        return ret;
+    }
+};