LeetCode，《剑指offer》中的算法题的题目和解法以及常见算法的实现
License: MIT License
判断奇偶性不对吧，应该是&1
bool isOdd(int val){
if (val & 2 == 1){
return true;
} else{
return false;
}
}
斐波那契数列不是从1 开始吗?第一个程序的代码判断是不是有问题呢
int result[2] = {0, 1};
if(n < 2){
return result[n];
}
传入0 返回 0(手动狗头)
这个如果链表的第一个节点和第二个节点相同，代码里第一个节点没有移除，比如1->1->2.
应该为Top Kth Frequent (出现频率最高的前k个元素)
根据实现，例子中N只能取到4，根据题目N可以最多取到5。
因此for循环应该从1开始到<=N
冒泡排序，每排一次，最外层的循环可以少一次，不用每次都i < n - 1，除非n--
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