算法分析与设计第一次作业

204.Count Primes

Count the number of prime numbers less than a non-negative number, n.

int countPrimes(int n) {

        if(n<=2) return 0;int i,j,k= 0;int count = 0;int flag = 0;int t = 1;int a[200];//用于保存已知的素数，减少循环次数a[0] = 2;for(i =3;i<n;i++) {for(j =0;j <=k;j++) {if(i%a[j] == 0) {flag = 1;break;}}if(flag == 0) {if(k<199) {k++;a[k] = i;}count ++;} else flag = 0;}return (count+1);}

104. Maximum Depth of Binary Tree

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

 int max(int x, int y){       return ((x > y ) ? x : y);}int maxDepth(struct TreeNode* root) {    if(root == NULL){        return 0;    }    int left = maxDepth(root->left);//递归的方法求树高，思路简单，效率较低    int right = maxDepth(root->right);    return 1 + max(left,right);}

121. Best Time to Buy and Sell Stock

Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

int maxProfit(int* prices, int pricesSize) {    int maxDiff = 0, curMin = prices[0];    for(int i = 1;i < pricesSize ; i++) {        maxDiff = maxDiff > (prices[i] - curMin) ? maxDiff : (prices[i] - curMin);//贪心算法，随时更新最小值和当前最大利润        curMin = curMin < prices[i] ? curMin : prices[i];    }    return maxDiff;}