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微軟面試一百道題目精選-wenkub.com

2025-03-22 01:59 本頁面
   

【正文】 if (testShares(a, n, m, sum, sum/m, aux, sum/m, 1)) return m。 i++) aux[i] = 0。 for (m=n。 for (i=0。 int sum = 0。 2: standard maximum flow. If the size of matrix is NxN, then the algorithm using Ford Fulkerson algorithm is M*N*N.too plex... I will do this when I have time...,長(zhǎng)度為n,將其分為m 份,使各份的和相等,求m 的最大值比如{3,2,4,3,6} 可以分成{3,2,4,3,6} m=1。請(qǐng)用5 分鐘時(shí)間,找出重復(fù)出現(xiàn)最多的前10 條。 break。 Y : { transformY(level)。 switch direction: { // 6 faces, 9 chips each face visit()。 } }}Postorder nonrecursive traversal is the hardest one. However, a simple observation helps that the node first traversed is the node last visited. This recalls the feature of stack. So we could use a stack to store all the nodes then pop them out altogether.This is a very elegant solution, while takes O(n) space. (current)。 while (!() || current != NULL) { stackTreeNode * s。 (node)。 preorderRecursive(noderight)。 visit(node)。}。 } candi = h1。amp。 Node * p = head。 head = h2。ANSWERI don’t quite understand what it means by “not modifying linked list’s data”. If some nodes will be given up, it is weird for this requirement.Node * head(Node *h1, Node * h2) {s memory very fast, which should be the result of execution of about a few thousands of simple statements. }}3)設(shè)計(jì)一個(gè)系統(tǒng)處理詞語搭配問題,比如說中國(guó)和人民可以搭配,則中國(guó)人民人民中國(guó)都有效。 if (min ht) min = ht。 } if (c[a[t]] == 0) break。amp。ANSWERUse a sliding window and a counting array, plus a counter which monitors the num of zero slots in counting array. When there is still zero slot(s), advance the window head, until there is no zero slot. Then shrink the window until a slot es zero. Then one candidate segment of (window_size + 1) is achieved. Repeat this. It is O(n) algorithm since each item is swallowed and left behind only once, and either operation is in constant time.int shortestFullcolor(int a[], int n, int m) { return low。 if (s == root amp。 } if (low visit[g[s][i]]) { if (visited[g[s][i]] == 0) { i=g[s][0]。 dfs(g, cuts, n, i, i)。 for (int i=0。 memset(visited, 0, sizeof(int)*n)。int lowest[MAX_NUM]。ANSWERThat is, keep total number count N. If N=m, just keep it.For Nm, generate a random number R=rand(N) in [0, N), replace a[R] with new number if R falls in [0, m). 字符串,如何從中去除重復(fù)的,優(yōu)化時(shí)間空間復(fù)雜度ANSWER1. Use hash map if there is enough memory.2. If there is no enough memory, use hash to put urls to bins, and do it until we can fit the bin into memory.39.網(wǎng)易有道筆試:(1).求一個(gè)二叉樹中任意兩個(gè)節(jié)點(diǎn)間的最大距離,兩個(gè)節(jié)點(diǎn)的距離的定義是這兩個(gè)節(jié)點(diǎn)間邊的個(gè)數(shù),比如某個(gè)孩子節(jié)點(diǎn)和父節(jié)點(diǎn)間的距離是1,和相鄰兄弟節(jié)點(diǎn)間的距離是2,優(yōu)化時(shí)間空間復(fù)雜度。 return 1。 finished[s] = 1。 if (visited[graph[s][i]] == 0) { return 0。 for (int i=1。 return 1。 for (int i=0。 memset(visit, 0, n*sizeof(int))。}int top[MAX_NUM]。 j++) { //topological sort failed, there is cycle.int longestPath(int graph[], int n) {*/int visit[MAX_NUM]。 return longestPath(graph, n)。 graph[i][0] ++。 // there is a self loop, return false. in。 } }}37.有n 個(gè)長(zhǎng)為m+1 的字符串,如果某個(gè)字符串的最后m 個(gè)字符與某個(gè)字符串的前m 個(gè)字符匹配,則兩個(gè)字符串可以聯(lián)接,問這n 個(gè)字符串最多可以連成一個(gè)多長(zhǎng)的字符串,如果出現(xiàn)循環(huán),則返回錯(cuò)誤。 for (i=0,j=0。 int i,j。 int round = n。n1,已知它們之間的實(shí)力對(duì)比關(guān)系,存儲(chǔ)在一個(gè)二維數(shù)組w[n][n]中,w[i][j] 的值代表編號(hào)為i,j 的隊(duì)伍中更強(qiáng)的一支。 jn。 i++) { i++) { acc[i] = a[i]。(2)分析時(shí)間復(fù)雜度。ANSWERIf only one swap can be taken, it is a O(n^2) searching problem, which can be reduced to O(nlogn) by sorting the arrays and doing binary search.If any times of swaps can be performed, this is a double binatorial problem.In the book beauty of codes, a similar problem splits an array to halves as even as possible. It is possible to take binary search, when SUM of the array is not too high. Else this is a quite time consuming brute force problem. I cannot figure out a reasonable solution.33.實(shí)現(xiàn)一個(gè)挺高級(jí)的字符匹配算法:給一串很長(zhǎng)字符串,要求找到符合要求的字符串,例如目的串:1231******3***2 ,12*****3 這些都要找出來其實(shí)就是類似一些和諧系統(tǒng)。 return count。 for (int j=0。 int count = 0。 //10 is enough for 32bit integers分析:這是一道廣為流傳的google 面試題。 i2++。return 0。 int i1=0, i2=0。比如輸入的push 序列是5,那么1 就有可能是一個(gè)pop 系列。 return c。 n = n amp。int countOf1(int n) { (xxxxxx100001) = xxxxxx00000Note: for negative numbers, this also hold, even with 100000000 where the “1” leading to an underflow.int countOf1(int n) {例如輸入10,由于其二進(jìn)制表示為1010,有兩個(gè)1,因此輸出2。求總共有多少總跳法,并分析算法的時(shí)間復(fù)雜度。如把字符串a(chǎn)bcdef 左旋轉(zhuǎn)2 位得到字符串cdefab。 *outputstr = ‘\0’。 } } else { len ++。 int max = 0。 int len = 0。 } current = currentnext。 h2=h2next。 if (h1 == NULL || (h2!=NULL amp。amp。 } h2=h2next。 if (h1datah2data) { if (h2 == NULL) return h1。3D 坐標(biāo)系原點(diǎn)(,)圓形:半徑r = 圓心o = (*.*, , *.*)正方形:4 個(gè)角坐標(biāo)。 printf(“\n”)。 for (int i=0。 helper(destidx1, idx+1, aux, n)。 helper(dest, idx+1, aux, n)。 dump(aux, n)。 if (dest == 0) memset(aux, 0, n*sizeof(int))。 if (nm) findCombination(m, m)。 }939。039。 while (*p != ‘\0’) { } else if (*p == ‘+’) { char * p = str。例如輸入字符串345,則輸出整數(shù)345。 multiply(tmp, A, _r)。 if (n amp。 power(A, n1, _r)。 return。分析:在很多C 語言教科書中講到遞歸函數(shù)的時(shí)候,都會(huì)用Fibonacci 作為例子。 }ANSWER:Actually, although this is a so traditional problem, I was always to lazy to think about this or even to search for the answer.(What a shame...). Finally, by google I found the elegant solution for it.The keys are:1) if we shift the ids by k, namely, start from k instead of 0, we should add the result by k%n2) after the first round, we start from k+1 ( possibly % n) with n1 elements, that is equal to an (n1) problem while start from (k+1)th element instead of 0, so the answer is (f(n1, m)+k+1)%n3) k = m1, so f(n,m)=(f(n1,m)+m)%n. // this must the one that occurs exact 1 time.}第18 題:題目:n 個(gè)數(shù)字(0,1,…,n1)形成一個(gè)圓圈,從數(shù)字0 開始,每次從這個(gè)圓圈中刪除第m 個(gè)數(shù)字(第一個(gè)為當(dāng)前數(shù)字本身,第二個(gè)為當(dāng)前數(shù)字的下一個(gè)數(shù)字)。 if (a[*p] == 1) return *p。 p++。 a[*p] ++。如輸入abaccdeff,則輸出b。 } else { (
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