微软2014实习生电话一面总结_sqrt helper(n)-优快云博客

本文解析了三道经典算法题目，包括利用二分法求正整数的平方根、判断两个链表是否相交以及求解二叉树的最大子树和。通过对这些题目进行深入分析，展示了算法设计的基本思想和实现技巧。

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几道算法题：

第一题：

// MS.cpp : 定义控制台应用程序的入口点。
//
#include <iostream>
using namespace std;
/*函数功能：求正整数的平方根，要求误差在0.001
 思路：二分法
 */

//在误差范围内比较两个数
int dcmp(double x) {
	if (x < -1e-3)
		return -1;
	return x > 1e-3;
}
double sqrtHelper(double left, double right, double n) {
	double mid = (left + right) / 2;
	if (dcmp(mid*mid - n) == 0)
		return mid;
	else if (dcmp(mid*mid - n) < 0)
		return sqrtHelper(mid, right, n);
	else
		return sqrtHelper(left, mid, n);
}

double mySqrt(double n) {
	return sqrtHelper(0, n, n);
}

int main() {
	int a = 5, b = 3, c = 9;
	double as = mySqrt(a);
	double bs = mySqrt(b);
	double cs = mySqrt(c);
	cout << as << endl;
	cout << bs << endl;
	cout << cs << endl;
	return 0;
}

第二题：

#include <iostream>
using namespace std;

/*函数功能：判断两个链表是否相交
 思路：考虑几种情况：1.两个链表都不带环 2.其中一个带环（定不相交）3.两个都带环
*/

struct LinkNode {
	int val;
	LinkNode *next;
	LinkNode(int x) : val(x), next(NULL){}
};
//两个链表都没有环，看尾指针是否相等
int isSimpleJoined(LinkNode *h1, LinkNode *h2) {
	LinkNode *p1 = h1, *p2 = h2;
	while (p1->next != NULL)
		p1 = p1->next;
	while (p2->next != NULL)
		p2 = p2->next;
	return p1 == p2;
}

//使用快慢指针判断链表是否有环
LinkNode *testCycle(LinkNode *h1) {
	LinkNode *slow = h1, *fast = h1;
	while (fast != NULL && fast->next != NULL) {
		slow = slow->next;
		fast = fast->next->next;
		if (slow == fast)
			return slow;
	}
	return NULL;
}

int isJoined(LinkNode *h1, LinkNode *h2) {
	LinkNode *c1 = testCycle(h1);
	LinkNode *c2 = testCycle(h2);
	//两链表都无环
	if (c1 == NULL && c2 == NULL)
		return isSimpleJoined(h1, h2);
	//只有一个链表有环
	else if (c1 == NULL || c2 == NULL)
		return 0;
	//两个链表都有环，c1,c2都在环中，若c1绕环一圈没遇到c2则两个链表不相交
	LinkNode *p1 = c1;
	while (1) {
		if (c1 == c2)
			return 1;
		p1 = p1->next->next;
		c1 = c1->next;
		if (p1 == c1)
			return 0;
	}
}

int main() {
	LinkNode *h1 = new LinkNode(1);
	h1->next = new LinkNode(2);
	h1->next->next = new LinkNode(3);
	h1->next->next->next = new LinkNode(4);
	h1->next->next->next->next = h1->next->next->next;
	LinkNode *h2 = new LinkNode(5);
	h2->next = new LinkNode(6);
	h2->next->next = h1->next->next;
	int join = isJoined(h1, h2);
	cout << join << endl;
	return 0;
}

第三题：

#include <iostream>
#include <limits.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;


struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {};
};
/*函数功能：求二叉树的最大子树和
*/

int maxS = INT_MIN;

int maxTreeSum(TreeNode *root) {
	if (root == NULL)
		return 0;
	int left = maxTreeSum(root->left);
	int right = maxTreeSum(root->right);
	int sum = left + right + root->val;
	maxS = max(sum, maxS);
	return sum;
}

int main() {
	TreeNode *root = new TreeNode(10);
	root->left = new TreeNode(-20);
	root->right = new TreeNode(5);
	root->left->left = new TreeNode(-7);
	root->left->right = new TreeNode(8);
	root->right->left = new TreeNode(7);
	root->right->right = new TreeNode(9);
	maxTreeSum(root);
	cout << maxS << endl;
	return 0;
}