题目一
在一个二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。
C++代码如下（暴力解法）：
class Solution {
public:
bool Find(int target, vector<vector<int> > array) {
bool isFound = false;
for(int i = 0; i < array.size(); i++){
for(int j = 0; j < array[i].size(); j++){
if(false == isFound && target == array[i][j]){
isFound = true;
}
}
}
return isFound;
}
};
JAVA代码如下（暴力解法）：
public class Solution {
public boolean Find(int target, int [][] array) {
boolean Flag = false;
for(int i = 0; i < array.length; i++){
for(int j = 0; j < array[i].length; j++) {
if(Flag == false && array[i][j] == target)
Flag = true;
}
}
return Flag;
}
C++最优解如下：
class Solution {
public:
bool Find(int target, vector<vector<int> > array) {
bool isFound = false;
int row = 0;
int col = array[0].size() - 1;
while (row < array.size() && col >= 0) {
if(target > array[row][col]) {
row++;
}else if(target < array[row][col]) {
col--;
}else {
isFound = true;
break;
}
}
return isFound;
}
};
JAVA最优解如下：
public class Solution {
public boolean Find(int target, int [][] array) {
boolean Flag = false;
int row = 0;
int col = array[0].length - 1;
while (row < array.length && col >= 0) {
if(target > array[row][col]) {
row++;
}else if(target < array[row][col]) {
col--;
}else {
Flag = true;
break;
}
}
return Flag;
}
}
题目二
请实现一个函数，将一个字符串中的空格替换成“%20”。例如，当字符串为We Are Happy.则经过替换之后的字符串为We%20Are%20Happy。
C++暴力解如下：
class Solution {
public:
void replaceSpace(char *str,int length) {
if (str == NULL || length <= 0)
return;
int i = length - 1, j;
int len = length;
//从后往前进行遍历
for (int i = length - 1; i >= 0; i--) {
if (str[i] == ' ') {
len += 2;
for (j = len - 1; j > i+2; j--) {
str[j] = str[j - 2];
}
str[i] = '%';
str[i + 1] = '2';
str[i + 2] = '0';
}
}
str[len] = ' ';
}
};
C++最优解如下：
class Solution {
public:
void replaceSpace(char *str,int length) {
if (str == nullptr && length < 0)
return;
//获得原始字符串长度和空格数
int OrignalLength = 0;
int NumberOfBlank = 0;
int i = 0;
while (str[i] != ' ') {
OrignalLength++;
if (str[i] == ' ')
NumberOfBlank++;
++i;
}
//将字符串空格替换为%20
int NewLength = OrignalLength + 2 * NumberOfBlank;
if (NewLength > length)
return;
int indexOfOriginal = OrignalLength;
int indexOfNew = NewLength;
while (indexOfOriginal >= 0 && indexOfNew > indexOfOriginal) {
if (str[indexOfOriginal] == ' ') {
str[indexOfNew--] = '0';
str[indexOfNew--] = '2';
str[indexOfNew--] = '%';
indexOfOriginal--;
}else {
str[indexOfNew--] = str[indexOfOriginal--];
}
}
}
};
JAVA最优解如下：
public class Solution {
public String replaceSpace(StringBuffer str) {
if(str == null)
return null;
StringBuilder sb = new StringBuilder();
for(int i = 0; i < str.length(); i++) {
if(String.valueOf(str.charAt(i)).equals(" ")) {
sb.append("%20");
}else {
sb.append(str.charAt(i));
}
}
return String.valueOf(sb);
}
}
题目三
输入一个链表，从尾到头打印链表每个节点的值。
C++最优解如下：
//利用栈和容器逆向打印链表
vector<int> printListFromTailToHead04(ListNode* head) {
vector <int> nodeVector;
stack<int> stack;
ListNode *p = head;
if (head != nullptr) {
while (p != nullptr) {
stack.push(p->val);
p = p->next;
}
while (!stack.empty()) {
nodeVector.push_back(stack.top());
stack.pop();
}
}
return nodeVector;
}
JAVA最优解如下：
import java.util.ArrayList;
import java.util.Stack;
public class Solution {
public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
Stack<Integer> stack = new Stack<Integer>();
while(listNode != null) {
stack.push(listNode.val);
listNode = listNode.next;
}
ArrayList<Integer> list = new ArrayList<Integer>();
while(!stack.empty()) {
list.add(stack.pop());
}
return list;
}
}
题目四
输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。
C++最优解如下：
TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> vin) {
int inLength = vin.size();
if (inLength == 0)
return NULL;
vector<int> left_pre, right_pre, left_in, right_in;
//创建根节点，根节点是前序遍历的第一个数
TreeNode* head = new TreeNode(pre[0]);
//找到中序遍历根节点所在位置，存放在变量rootIndex中
int rootIndex = 0;
for (int i = 0; i < inLength; i++) {
if (vin[i] == pre[0]) {
rootIndex = i;
break;
}
}
//对于中序遍历，根节点左边的节点位于二叉树的左边，根节点右节点位于二叉树的右边
//利用上述这点，对二叉树进行归并
for (int i = 0; i < rootIndex; i++) {
left_in.push_back(vin[i]);
left_pre.push_back(pre[i+1]);//前序第一个为根节点
}
for (int i = rootIndex + 1; i < inLength; i++) {
right_in.push_back(vin[i]);
right_pre.push_back(pre[i]);
}
//取出前序和中序遍历根节点左子树和右字树，进行递归再区分左右字树直到叶节点
head->left = reConstructBinaryTree(left_pre, left_in);
head->right = reConstructBinaryTree(right_pre, right_in);
return head;
}
JAVA最优解如下：
//已知前序和中序遍历，重建二叉树返回头结点
public static TreeNode reConstructBinaryTree(int [] pre,int [] in) {
TreeNode root = reConstructBinaryTree(pre, 0, pre.length-1, in, 0, in.length-1);
return root;
}
private static TreeNode reConstructBinaryTree(int[] pre,int startPre,int endPre,int[] in,int startIn,int endIn) {
if(startPre > endPre || startIn > endIn)
return null;
TreeNode root = new TreeNode(pre[startPre]);
for(int i = startIn; i <= endIn; i++) {
if(in[i] == pre[startPre]) {
root.left = reConstructBinaryTree(pre, startPre+1, startPre+i-startIn, in, startIn, i-1);
root.right = reConstructBinaryTree(pre, i-startIn+startPre+1, endPre, in, i+1, endIn);
break;
}
}
return root;
}
题目五
用两个栈来实现一个队列，完成队列的Push和Pop操作。 队列中的元素为int类型。
C++最优解如下：
class Solution
{
public:
void push(int node) {
stack1.push(node);
}
int pop() {
if (stack2.empty()) {
while (!stack1.empty()) {
int value = stack1.top();
stack2.push(value);
stack1.pop();
}
}
int value = stack2.top();
stack2.pop();
return value;
}
private:
stack<int> stack1;
stack<int> stack2;
};
JAVA最优解如下：
import java.util.Stack;
public class Solution {
Stack<Integer> stack1 = new Stack<Integer>();
Stack<Integer> stack2 = new Stack<Integer>();
public void push(int node) {
stack1.push(node);
}
public int pop() {
if(stack2.empty()) {
while(!stack1.empty()) {
stack2.push(stack1.pop());
}
}
return stack2.pop();
}
}
题目六
把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。 输入一个非递减排序的数组的一个旋转，输出旋转数组的最小元素。 例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1。 NOTE：给出的所有元素都大于0，若数组大小为0，请返回0。
C++暴力解如下：
class Solution {
public:
int minNumberInRotateArray(vector<int> rotateArray) {
int length = rotateArray.size();
if (length == 0)
return 0;
for (int i = 0; i <= length-1; i++) {
if (rotateArray[i + 1] < rotateArray[i]) {
return rotateArray[i + 1];
}
}
return rotateArray[0];
}
};
JAVA暴力解如下：
import java.util.ArrayList;
public class Solution {
public int minNumberInRotateArray(int[] array) {
if (array.length == 0)
return 0;
for (int i = 0; i < array.length - 1; i++) {
if (array[i] > array[i + 1])
return array[i + 1];
}
return array[0];
}
}
C++最优解如下：
class Solution {
public:
int minNumberInRotateArray(vector<int> rotateArray) {
int low = 0 ; int high = rotateArray.size() - 1;
while(low < high){
int mid = low + (high - low) / 2;
if(rotateArray[mid] > rotateArray[high]){
low = mid + 1;
}else if(rotateArray[mid] == rotateArray[high]){
high = high - 1;
}else{
high = mid;
}
}
return rotateArray[low];
}
};
JAVA最优解如下：
import java.util.ArrayList;
public class Solution {
public int minNumberInRotateArray(int[] array) {
if (array.length == 0)
return 0;
int left = 0;
int right = array.length - 1;
int middle = -1;
while (array[left]>=array[right]) {
if(right-left==1){
middle = right;
break;
}
middle = left + (right - left) / 2;
if (array[middle] >= array[left]) {
left = middle;
}
if (array[middle] <= array[right]) {
right = middle;
}
}
return array[middle];
}
}
题目七
大家都知道斐波那契数列，现在要求输入一个整数n，请你输出斐波那契数列的第n项。n<=39
C++最优解如下：
class Solution {
public:
int Fibonacci(int n) {
int result[2] = {0,1};
if (n < 2) {
return result[n];
}
long long fibNMinusOne = 1;
long long fibNMinusTwo = 0;
long long fibN = 0;
for (int i = 2; i <= n; i++) {
fibN = fibNMinusOne + fibNMinusTwo;
fibNMinusTwo = fibNMinusOne;
fibNMinusOne = fibN;
}
return fibN;
}
};
JAVA最优解如下：
public class Solution {
public int Fibonacci(int n) {
int[] result = {0,1};
if(n < 2) {
return result[n];
}
int FiboNMinusOne = 1;
int FiboNMinusTwo = 0;
int FiboN = 0;
for(int i=2; i <= n; i++) {
FiboN = FiboNMinusOne + FiboNMinusTwo;
FiboNMinusTwo = FiboNMinusOne;
FiboNMinusOne = FiboN;
}
return FiboN;
}
}
题目八
一只青蛙一次可以跳上1级台阶，也可以跳上2级。求该青蛙跳上一个n级的台阶总共有多少种跳法。
C++最优解如下：
class Solution {
public:
int jumpFloor(int number) {
int result[2] = {1,2};
if(number <= 2)
return result[number-1];
int FiboNMinusOne = 2;
int FiboNMinusTwo = 1;
int FiboN = 0;
for(int i=3; i <= number; i++) {
FiboN = FiboNMinusOne + FiboNMinusTwo;
FiboNMinusTwo = FiboNMinusOne;
FiboNMinusOne = FiboN;
}
return FiboN;
}
};
JAVA最优解如下：
public class Solution {
public int JumpFloor(int target) {
int[] result = {1,2};
if(target <= 2)
return result[target-1];
int FiboNMinusOne = 2;
int FiboNMinusTwo = 1;
int FiboN = 0;
for(int i=3; i <= target; i++) {
FiboN = FiboNMinusOne + FiboNMinusTwo;
FiboNMinusTwo = FiboNMinusOne;
FiboNMinusOne = FiboN;
}
return FiboN;
}
}
题目九
一只青蛙一次可以跳上1级台阶，也可以跳上2级……它也可以跳上n级。求该青蛙跳上一个n级的台阶总共有多少种跳法。
C++最优解如下：
class Solution {
public:
int jumpFloorII(int number) {
int result[2] = {0,1};
if(number == 1 || number == 0)
return result[number];
int initialNum = 1;
int jumpNum = 0;
for(int i=2; i <= number; i++){
jumpNum = 2 * initialNum;
initialNum = jumpNum;
}
return jumpNum;
}
};
JAVA最优解如下：
public class Solution {
public int JumpFloorII(int target) {
int[] result = {0,1};
if(target == 1 || target == 0)
return result[target];
int initialNum = 1;
int jumpNum = 0;
for(int i=2; i <= target; i++){
jumpNum = 2 * initialNum;
initialNum = jumpNum;
}
return jumpNum;
}
}
题目十
我们可以用2*1的小矩形横着或者竖着去覆盖更大的矩形。请问用n个2*1的小矩形无重叠地覆盖一个2*n的大矩形，总共有多少种方法？
C++最优解如下：
class Solution {
public:
int rectCover(int number) {
int result[2] = {1,2};
if(number == 1 || number == 2)
return result[number-1];
int fiboNMinusOne = 2;
int fiboNMinusTwo = 1;
int fibN = 0;
for(int i=3; i<=number; i++){
fibN = fiboNMinusOne + fiboNMinusTwo;
fiboNMinusTwo = fiboNMinusOne;
fiboNMinusOne = fibN;
}
return fibN;
}
};
JAVA最优解如下：
public class Solution {
public int RectCover(int target) {
int[] result = {1,2};
if(target == 1 || target == 2)
return result[target-1];
int fiboNMinusOne = 2;
int fiboNMinusTwo = 1;
int fibN = 0;
for(int i=3; i<=target; i++){
fibN = fiboNMinusOne + fiboNMinusTwo;
fiboNMinusTwo = fiboNMinusOne;
fiboNMinusOne = fibN;
}
return fibN;
}
}