今日头条 面试题一例

之前同学面试今日头条，碰到了破浪数问题
即在数组上升下降不定时，找到其中一个波峰
如{1,2,3, 1,0,-1, -3,1,3, 4,5,2, 1,0,2, 3,2,1}找到3即可

最优为O(logn)解法
随手写了一下，马马虎虎能满足要求；望指正

public class test{
    // static int[] array = new int[]{1,2,3, 1,0,-1, -3,1,3, 4,5,2, 1,0,2, 3,2,1};
    // static int[] array = new int[]{1,2,3, 1,0,-1, -3,1};
    static int[] array = new int[10];
    public static int inner(int left, int right ){
        System.out.println("left"+left+"right"+right);
        if (left<=0 || right>=array.length-1 || right-left<0){
            return -32767;
        }

        if (left == right || right-left == 1){
            if(array[left]-array[left-1]>0 && array[right+1]-array[right]<0){
                return Math.max(array[left], array[right]);
            }else if (array[left]-array[left-1]==0 && array[right+1]-array[right]==0){
                return Math.max(array[left], array[right]);
            }
            else{return -32767;}

        }
        if(array[left]-array[left-1]>0 && array[right+1]-array[right]<0){
            right = left+(right-left+1)/2 ;
            return inner(left, right);
        }else if(array[right]-array[left-1]>0 && array[right+1]-array[right]>0){
            int temp = right-left;
            int returnValue = inner(right, right+temp);
            if(returnValue == -32767){
                returnValue = inner(left+1, right);
            }
            return returnValue;

        }else if(array[left]-array[left-1]<0 && array[right+1]-array[right]<0){
            int temp = right-left;

            int returnValue = inner(left-temp, left);
            if(returnValue == -32767){
                returnValue = inner(left, right-1);
            }
            return returnValue;
            // right++;
        }
        else if(array[left]-array[left-1]<0 && array[right+1]-array[right]>0){
            right--;
            // left++;
        }else{
            right--;
            // left++;
        }
        return inner(left, right);
    }

    public static int fourth(){

    }
    public static void main(String[] args) {
        for(int i : array)
        System.out.println(i);
        // O(log n)
        // for(int i=0;i
        int left = 1;
        int right = array.length-2;
        int result = 0;
        result = inner(left, right);
        System.out.println(result);
    }
}

还有一道题是判断是否是平衡二叉树的
如果某二叉树中任意结点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。
源于剑指offer：

    bool IsBalanced(BinaryTreeNode* pRoot, int* depth)  
    {  
        if(pRoot== NULL)  
        {  
            *depth = 0;  
            return true;  
        }  

        int nLeftDepth,nRightDepth;  
        bool bLeft= IsBalanced(pRoot->m_pLeft, &nLeftDepth);  
        bool bRight = IsBalanced(pRoot->m_pRight, &nRightDepth);  

        if (bLeft && bRight)  
        {  
            int diff = nRightDepth-nLeftDepth;  
            if (diff<=1 || diff>=-1)  
            {  
                *depth = 1+(nLeftDepth > nRightDepth ? nLeftDepth : nRightDepth);  
                return true;  
            }  
        }  

        return false;  
    }  

    bool IsBalanced(BinaryTreeNode* pRoot)  
    {  
        int depth = 0;  

        return IsBalanced(pRoot, &depth);  
    }