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《剑指Offer》、LeetCode

Java 99.87% Lua 0.13%

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剑指offer 18 题。

题目：输入两棵二叉树 A 和 B，判断 B 是不是 A 的子结构。
【解1】：

 利用二叉树的神级遍历(空间复杂度：O(1)， 时间复杂度：O(h))

 转化成字符串匹配问题：KMP算法（时间：O(N)）

```
 总的时间复杂度：O(N)
```

看起来是一种解有bug 如果通过先序遍历不能够确定一个树把。就是说加入两棵树结构不一样，也能返回相同字符串吧。

a_Selection 选择排序算法

插入排序算法
public static int[] Selection_Sort(int[] a) {
```
    int N = a.length;
```
```
    int min = 0;
```
```
    for (int i = 0; i < N - 1; i++) {
```
```
    	min = i;
```

        for (int j = i + 1; j < N; j++) {

```
            if (a[min] > a[j]) {
```
```
                min = j;
```
```
            }
```
```
        }
```
```
        std.swap(a, i, min);
```
```
    }
```
```
    return a;
```
```
}
```
该方法第9行缺少min=i; 最小值回跳操作

剑指offer

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