一天一道leetcode，先ac，再比较最优解法
1.两数之和
class Solution {
public:
vector twoSum(vector& nums, int target) {
vector out;
for(int i = 0 ; i < nums.size() ; ++i){
int m = target - nums[i];
for(int j = i+1 ; j < nums.size(); ++j){
if(nums[j]==m){
out.push_back(i);
out.push_back(j);
break;
}
}
}
return out;
}
};
执行用时: 192 ms, 在Two Sum的C++提交中击败了10.58% 的用户
2.两数相加（未ac）
class Solution{
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* l;
for(int i = 0 ; ; ++i){
while(l1!=NULL || l2!=NULL){
int yu = 0;
int x = l1->val;
int y = l2->val;
if((x+y+yu)>9){
l->val = (x+y+yu)-10;
l=l->next;
yu = 0;
l1 = l1->next;
l2 = l2->next;
}
else{
l->val = x+y;
l=l->next;
l1 = l1->next;
l2 = l2->next;
}
}
return l;
}
}
};
//修改后代码 AC
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode* l = NULL,**t = &l;
int yu = 0,x = 0, y = 0,sum = 0;
while(l1!=NULL || l2!=NULL){
if(l1 != NULL)x = l1->val;else x = 0;
if(l2 != NULL)y = l2->val;else y = 0;
sum = x+y+yu;
ListNode *node = new ListNode(sum%10);
*t = node;
t=(&node->next);
yu = sum/10;
if(l1 != NULL)l1 = l1->next;
if(l2 != NULL)l2 = l2->next;
}
if(yu){
ListNode *node = new ListNode(yu%10);
*t = node;
}
return l;
}
};
//注意：链表数字可能出现不等长情况，用等长情况写的代码容易出现指向空指针，所以需要判断是否是尾链表，如果是尾链表还要注意以后的取值都为零
11.盛最多水的容器
class Solution {
public:
int maxArea(vector& height) {
int max = 0;
for(int i = 0 ; i < height.size()-1 ; i++){
for(int j = i+1 ; j < height.size() ; j++){
int h = (height[i]<height[j])? height[i]:height[j];
if(max < h*(j-i))max = h*(j-i);
}
}
return max;
}
};