此博客彙總劍指offer上的部分代碼，更新中.......
6.重建二叉樹（前序和中序）
TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> vin) {
if (pre.empty() || vin.empty())
return NULL;
return reConstruct(pre, vin, 0, pre.size() - 1, 0, vin.size() - 1);
}
TreeNode* reConstruct(vector<int> &pre, vector<int> &vin, int pStart, int pEnd, int vStart, int vEnd) {
if (vStart>vEnd)
return NULL;
TreeNode *root = new TreeNode(pre[pStart]);
for (int i = vStart; i <= vEnd; i++) {
if (pre[pStart] == vin[i]) {
root->left = reConstruct(pre, vin, pStart + 1, pStart + i - vStart, vStart, i - 1);
root->right = reConstruct(pre, vin, pEnd - vEnd + i + 1, pEnd, i + 1, vEnd);
break;
}
}
return root;
}17.合併兩個排序鏈表
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
if (!pHead1)
return pHead2;
if (!pHead2)
return pHead1;
if (pHead1->val<pHead2->val) {
pHead1->next = Merge(pHead1->next, pHead2);
return pHead1;
}
else {
pHead2->next = Merge(pHead1, pHead2->next);
return pHead2;
}
}bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2)
{
if (!pRoot1 || !pRoot2) return false;
return IsSubtree(pRoot1, pRoot2) || HasSubtree(pRoot1->left, pRoot2) || HasSubtree(pRoot1->right, pRoot2);
}
bool IsSubtree(TreeNode* pRoot1, TreeNode* pRoot2) {
if (!pRoot2) return true; //注意判斷的順序
if (!pRoot1) return false;
if (pRoot1->val != pRoot2->val) return false;
return IsSubtree(pRoot1->left, pRoot2->left) && IsSubtree(pRoot1->right, pRoot2->right);
}20.順時針打印矩陣
vector<int> spiralOrder(vector<vector<int>>& matrix) {
vector<int> res;
if (matrix.empty())
return res;
int m = matrix.size() - 1, n = matrix[0].size() - 1;
for (int i = 0, j = 0; i <= m && j <= n; i++, j++) {
for (int k = j; k <= n; k++)
res.push_back(matrix[i][k]);
for (int k = i + 1; k <= m; k++)
res.push_back(matrix[k][n]);
for (int k = n - 1; k >= j && i != m; k--)
res.push_back(matrix[m][k]);
for (int k = m - 1; k>i && j != n; k--)
res.push_back(matrix[k][j]);
m--;
n--;
}
return res;
}21.包含min函數的棧
class MinStack {
public:
int min = INT_MAX;
stack<int> stack1;
MinStack() {
}
void push(int x) {
if (x <= min) {
stack1.push(min);
min = x;
}
stack1.push(x);
}
void pop() {
if (stack1.top() == min) {
stack1.pop();
min = stack1.top();
}
stack1.pop();
}
int top() {
return stack1.top();
}
int getMin() {
return min;
}
};22.棧的輸入輸出隊列
bool IsPopOrder(vector<int> pushV, vector<int> popV) {
stack<int> stk;
int j = 0;
for (int i = 0; i < pushV.size(); i++) {
stk.push(pushV[i]);
while (j < popV.size() && stk.top() == popV[j]) {
stk.pop();
j++;
}
}
return stk.empty();
}27.二叉樹轉雙向鏈表
TreeNode* Convert(TreeNode* pRootOfTree)
{
if (pRootOfTree == NULL) return NULL;
pRootOfTree = ConvertNode(pRootOfTree);
while (pRootOfTree->left) pRootOfTree = pRootOfTree->left;
return pRootOfTree;
}
TreeNode* ConvertNode(TreeNode* root) {
if (root->left) {
TreeNode* left = ConvertNode(root->left);
while (left->right) left = left->right;
left->right = root;
root->left = left;
}
if (root->right) {
TreeNode* right = ConvertNode(root->right);
while (right->left) right = right->left;
right->left = root;
root->right = right;
}
return root;
}29.數組中出現次數超過一半的數字
int majorityElement(vector<int>& nums) {
int x = nums[0], count = 1;
for (int i = 1; i < nums.size(); i++) {
if (count == 0) {
x = nums[i];
count = 1;
continue;
}
if (nums[i] == x) count++;
else count--;
}
return x;
}int NumberOf1Between1AndN_Solution(int n)
{
int count = 0, loc = 1;
int high = 0, cur = 0, low = 0;
while (n / loc) {
low = n - (n / loc)*loc;
cur = (n / loc) % 10;
high = n / (loc * 10);
if (cur == 0)
count += high*loc;
else if (cur == 1)
count += high*loc + low + 1;
else
count += (high + 1)*loc;
loc = loc * 10;
}
return count;
}36.數組中逆序對
int result;
void merge(vector<int> &nums, int left, int right, int mid) {
vector<int> temp(right - left + 1, 0);
int lstart = left, rstart = mid + 1;
int tempIndex = 0;
while (lstart <= mid && rstart <= right) {
if (nums[lstart] <= nums[rstart])
temp[tempIndex++] = nums[rstart++];
else {
temp[tempIndex++] = nums[lstart++];
result += right - rstart + 1;
}
}
while (lstart <= mid) temp[tempIndex++] = nums[lstart++];
while (rstart <= right) temp[tempIndex++] = nums[rstart++];
for (int i = left; i <= right; i++) {
nums[i] = temp[i - left];
}
}
void mergeSort(vector<int> &nums, int left, int right) {
if (left < right) {
int mid = (left + right) >> 1;
mergeSort(nums, left, mid);
mergeSort(nums, mid + 1, right);
merge(nums, left, right, mid);
}
}39.平衡二叉樹判斷
bool IsBalanced(TreeNode* pRoot, int &depth) {
if (!pRoot) {
depth = -1;
return true;
}
int left, right;
if (IsBalanced(pRoot->left, left) && IsBalanced(pRoot->right, right)) {
int diff = left - right;
if (diff <= 1 && diff >= -1) {
depth = 1 + (left>right ? left : right);
return true;
}
}
return false;
}
bool IsBalanced_Solution(TreeNode* pRoot) {
int depth;
return IsBalanced(pRoot, depth);
}42.翻轉單詞順序
string ReverseSentence(string str) {
int len = str.size();
int start = 0;
for (int i = 0; i < len; i++)
{
if (str[i] == ' ')
{
reverse(str.begin() + start, str.begin() + i);
start = i + 1;
}
if (i == len - 1)
{
reverse(str.begin() + start, str.end());
}
}
reverse(str.begin(), str.end());
return str;
}47.不用加減乘除做加法
int Add(int num1, int num2){
if (!num2) return num1;
return Add(num1^num2, (num1&num2) << 1);
}49.把字符串轉化成整數
bool isTrue = false;
long long myAtoi(string str) {
int i = 0, minus = 1;
long long result = 0;
if (str.empty())
return result;
while (i < str.size()) {
if (str[i] == ' ') {
i++;
continue;
}
if (str[i] == '-') {
minus = -1;
i++;
}
else if (str[i] == '+')
i++;
break;
}
while (i < str.size()) {
if (str[i] >= '0' && str[i] <= '9') {
result = result * 10 + minus*(str[i] - '0');
if (minus == 1 && result >INT_MAX || minus == -1 && result < INT_MIN)
result = minus == 1 ? INT_MAX : INT_MIN;
i++;
}
else
break;
}
if (i == str.size())
isTrue = true;
return result;
}bool getPath(TreeNode* root, TreeNode* node, vector<TreeNode*> &path) {
if (root == node) {
path.push_back(root);
return true;
}
path.push_back(root);
if (root->left && getPath(root->left, node, path) ||
root->right && getPath(root->right, node, path))
return true;
path.pop_back();
return false;
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (!root || !p || !q) return NULL;
vector<TreeNode*> path1;
vector<TreeNode*> path2;
getPath(root, p, path1);
getPath(root, q, path2);
TreeNode *plast = NULL;
for (int i = 0; i<path1.size() && i <path2.size(); i++) {
if (path1[i] == path2[i])
plast = path1[i];
else
break;
}
return plast;
}bool duplicate(int numbers[], int length, int* duplication) {
if (numbers == NULL || length <= 0)
return false;
for (int i = 0; i<length; i++) {
if (numbers[i]<0 || numbers[i]>length - 1)
return false;
}
for (int i = 0; i<length; i++) {
while (i != numbers[i]) {
if (numbers[i] == numbers[numbers[i]]) {
*duplication = numbers[i];
return true;
}
swap(numbers[i], numbers[numbers[i]]);
}
}
return false;
}59.對稱的二叉樹
bool SymmetircFunc(TreeNode* node1, TreeNode *node2){
if (!node1 && !node2) return true;
if (!node1 || !node2) return false;
if (node1->val != node2->val) return false;
return SymmetircFunc(node1->right, node2->left) && SymmetircFunc(node1->left, node2->right);
}
bool isSymmetrical(TreeNode* pRoot){
if (!pRoot) return true;
return SymmetircFunc(pRoot->left, pRoot->right);
}64.數據流中的中位數
class Solution {
public:
void Insert(int num)
{
count++;
if (count & 0x1 == 1) {
minheap.push(num);
maxheap.push(minheap.top());
minheap.pop();
}
else {
maxheap.push(num);
minheap.push(maxheap.top());
maxheap.pop();
}
}
double GetMedian()
{
if (count & 0x1 == 1)
return maxheap.top();
else
return (maxheap.top() + minheap.top()) / 2.0; //注意浮點數
}
private:
int count = 0;
priority_queue<int> maxheap;
priority_queue<int, vector<int>, greater<int>> minheap;
};GitHub-Leetcode: https://github.com/wenwu313/LeetCode