1. 斐波那契數(shù)列求和
斐波那契數(shù)列（0掌桩，1边锁，1，2波岛，3茅坛，5 . . .）求和
- (NSInteger )fb:(NSInteger)n{
if (n <= 1) {
return n;
}
NSInteger first = 0;
NSInteger second = 1;
NSInteger sum = 0;
for (NSInteger i = 0; i < n-1; i++) {
sum = first + second;
first = second;
second = sum;
}
return second;
}
- 反轉(zhuǎn)鏈表(反轉(zhuǎn)鏈表)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public ListNode reverseList(ListNode head) {
ListNode preNode = null;
ListNode curNode = head;
while ( curNode != null ){
ListNode nextNode = curNode.next;
curNode.next = preNode;
preNode = curNode;
curNode = nextNode;
}
return preNode;
}
- 合并鏈表( 合并兩個(gè)有序鏈表)
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
public:
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if (l1 == NULL){
return l2;
}else if(l2 == NULL){
return l1;
}
ListNode *mergedNode = NULL;
if(l1->val < l2->val){
mergedNode = l1;
mergedNode->next = mergeTwoLists(mergedNode->next,l2);
}else{
mergedNode = l2;
mergedNode->next = mergeTwoLists(l1, mergedNode->next);
}
return mergedNode;
}
4.查找鏈表倒數(shù)第k個(gè)元素( 面試題 02.02. 返回倒數(shù)第 k 個(gè)節(jié)點(diǎn))
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
int kthToLast(ListNode* head, int k) {
if(head == NULL )
return -1;
ListNode *pAhead = head;
ListNode *pBehind = head;
for(int i = 0;i < k - 1;i++)
{
if(pAhead->next != NULL)
pAhead = pAhead->next;
else
{
return -1;
}
}
while(pAhead->next != NULL)
{
pAhead = pAhead->next;
pBehind = pBehind->next;
}
return pBehind->val;
}
};
5.倒序打印鏈表的值
struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
void PrintListReversingly(ListNode* pHead)
{
if(pHead != NULL)
{
if (pHead->next != NULL)
{
PrintListReversingly(pHead->next);
}
printf("%d\t", pHead->val);
}
}
6.刪除某個(gè)節(jié)點(diǎn)( 刪除鏈表中的節(jié)點(diǎn))
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
void deleteNode(struct ListNode* node) {
if(node == NULL || node->next == NULL){
return;
}
node->val = node->next->val;
node->next = node->next->next;
}
- 刪除值為val的節(jié)點(diǎn)
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeElements(ListNode *head, int val){
ListNode *dummyHead = new ListNode(-1);
dummyHead->next = head;
ListNode *cur = dummyHead;
while (cur->next != NULL){
if (cur->next->val == val){
ListNode *delNode = cur->next;
cur->next = delNode->next;
delete delNode;
} else{
cur = cur->next;
}
}
ListNode *returnNode = dummyHead->next;
delete dummyHead;
return returnNode;
}
};
- 二叉樹(shù)轉(zhuǎn)鏈表( 114. 二叉樹(shù)展開(kāi)為鏈表)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
TreeNode prev ;
public void flatten(TreeNode root) {
if (root == null)
return;
flatten(root.right);
flatten(root.left);
if (prev != null){
root.right = prev;
root.left = null;
}
prev = root;
}
}
64. 最小路徑和
int minPathSum(vector<vector<int>>& grid) {
int n = grid.size();
assert(n > 0);
int m = grid[0].size();
assert(m > 0);
vector<vector<int>> res = grid;
for (int i = 1;i < n;i++){
res[i][0] = grid[i][0] + res[i-1][0];
}
for (int i = 1;i < m;i++){
res[0][i] = grid[0][i] + res[0][i-1];
}
for (int i = 1;i<n;i++){
for (int j = 1;j<m;j++){
res[i][j] = min(res[i-1][j],res[i][j-1])+grid[i][j];
}
}
return res[n-1][m-1];
}
5. 最長(zhǎng)回文子串
string longestPalindrome(string s) {
if(s.size() <= 1) return s;
int maxLen = 1;
int begin = 0;
int i = 0;
while(i<s.size()){
int l = i-1;
int r = i;
while(++r < s.size() && s[r] == s[i]){
}
i = r;
while(l>=0 && r<s.size() && s[l]== s[r]){
l--;
r++;
}
int len = r-l-1;
if(maxLen < len){
maxLen = len;
begin = l+1;
}
}
return s.substr(begin,maxLen);
}
236. 二叉樹(shù)的最近公共祖先
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(root == NULL || root == p || root == q){
return root;
}
TreeNode *left = lowestCommonAncestor(root->left,p,q);
TreeNode *right = lowestCommonAncestor(root->right,p,q);
if(left != NULL && right != NULL){
return root;
}
return (left != NULL)?left:right;
}
