454. 四数相加 II
class Solution {
public:
int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
unordered_map<int,int>umap;
for(int i=0;i<nums1.size();i++){
for(int j=0;j<nums2.size();j++){
umap[nums1[i]+nums2[j]]++;
}
}
int vount=0;
for(int i=0;i<nums3.size();i++){
for(int j=0;j<nums4.size();j++){
if(umap.find(0-nums3[i]-nums4[j])!=umap.end()){
vount+=umap[0-nums3[i]-nums4[j]];
}
}
}
return vount;
}
};
383. 赎金信
没什么好写的,和之前的一个题差不多
class Solution {
public:
bool canConstruct(string ransomNote, string magazine) {
int record[26] = {
0};
//add
if (ransomNote.size() > magazine.size()) {
return false;
}
for (int i = 0; i < magazine.length(); i++) {
// 通过record数据记录 magazine里各个字符出现次数
record[magazine[i]-'a'] ++;
}
for (int j = 0; j < ransomNote.length(); j++) {
// 遍历ransomNote,在record里对应的字符个数做--操作
record[ransomNote[j]-'a']--;
// 如果小于零说明ransomNote里出现的字符,magazine没有
if(record[ransomNote[j]-'a'] < 0) {
return false;
}
}
return true;
}
};
第15题. 三数之和
逻辑不难,难的是有效去重
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> res;
sort(nums.begin(),nums.end());
for(int i=0;i<nums.size();i++){
if(nums[i]>0)return res;
if(i>0&&nums[i]==nums[i-1])continue;
int left=i+1;
int right=nums.size()-1;
while(left<right){
if(nums[i]+nums[left]+nums[right]>0){
right--;
}else if(nums[i]+nums[left]+nums[right]<0){
left++;
}else{
while (right > left && nums[right] == nums[right - 1]) right--;
while (right > left && nums[left] == nums[left + 1]) left++;
res.push_back({
nums[i],nums[left],nums[right]});
left++;
right--;
}
}
}
return res;
}
};
第18题. 四数之和
很难全做对
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
sort(nums.begin(),nums.end());
for(int i=0;i<nums.size();i++){
if(nums[i]>target&&nums[i] >= 0)return res;
if(i>0&&nums[i]==nums[i-1])continue;
for(int j=i+1;j<nums.size();j++){
if(nums[i]+nums[j]>target&&nums[i]+nums[j] >= 0)break;
if(j>i+1&&nums[j]==nums[j-1])continue;
int left=j+1;
int right=nums.size()-1;
while(left<right){
if((long)nums[i]+nums[left]+nums[right]+nums[j]>target){
right--;
}else if((long)nums[i]+nums[left]+nums[right]+nums[j]<target){
left++;
}else{
res.push_back({
nums[i],nums[left],nums[right],nums[j]});
while (right > left && nums[right] == nums[right - 1]) right--;
while (right > left && nums[left] == nums[left + 1]) left++;
left++;
right--;
}
}
}
}
return res;
}
};