1. 兩數之和
利用map容器“一對一”記錄兩數關係. 主要思路是查找對偶數other是否被記錄, 如果無,則將當前元素作爲對偶數存入map;如果存在對偶數,則返回二者的索引. 時間複雜度, 空間複雜度
class Solution
{
public:
vector<int> twoSum(vector<int>& nums, int target)
{
map<int, int> record;
for(int i=0; i<nums.size(); i++)
{
int other = target - nums[i];
if(record.count(other))
{
return {i, record[other]};
}
else
{
other = nums[i];
record[other] = i;
}
}
return {-1,-1};
}
};
2. 兩數相加
考慮創建啞節點dummy,使用dummy->next表示真正的頭節點,避免空邊界. 時間複雜度
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* first = new ListNode(0); // 啞結點
ListNode* last = first;
int carry = 0;
while(l1 or l2 or carry){
// 相加
int bitsum = 0;
if(l1){
bitsum += l1->val;
l1 = l1->next;
}
if(l2){
bitsum += l2->val;
l2 = l2->next;
}
if(carry){
bitsum += carry;
}
// 結果
int digit = bitsum % 10;
carry = bitsum / 10;
// 鏈表存儲
ListNode* node = new ListNode(digit);
last->next = node;
last = node;
}
last = first->next;
delete first;
return last;
}
};
3. 無重複字符的最長子串
以滑動窗口的方式來尋找子串, 左指針在遇到重複元素時更新, 右指針即遍歷指針i. 時間複雜度
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
length = l = r = 0
while r<len(s):
if s[r] not in s[l:r]:
r += 1
length = max(length, r-l)
else:
l += 1
return length
用python做算法實現,效率不高,但是很好理解。