day46
139.单词拆分
题目链接:139. 单词拆分
回溯暴搜
// 回溯暴搜超时
class Solution {
public:
bool backtracking( const string& s,const unordered_set<string>& wordset,int startIndex){
if( startIndex >= s.size()){
return true;
}
for( int i = startIndex;i < s.size(); i++){
string word = s.substr(startIndex,i-startIndex+1);
if( wordset.find(word) != wordset.end() && backtracking(s,wordset,i+1)){
return true;
}
}
return false;
}
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> wordset(wordDict.begin(),wordDict.end());
return backtracking( s,wordset,0);
}
};
"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]
动态规划
单词是物品,字符串s是背包
动态规划五部曲:
1.确定dp[j]下标及含义: dp[j]表示字符串长度为j的话,dp[j]为true
2.确定递推公式:如果dp[i]为true,且[i,j]区间的子串出现在这个字典中,dp[j]一定时true(i<j).
3.初始化:dp[0] = true,剩下都是false;
4.确定遍历顺序:本题相当于排列问题,先背包再物品
for( int j = 0; j <= s.size(); j++ ){ for( int i = 0; i < j; i++){ string word = s.substr(i,j-i+1); if( wordset.find(word) != wordset.end() && dp[i]){ dp[j] = true; } } }
5.打印dp数组
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
unordered_set<string> wordset(wordDict.begin(),wordDict.end());
vector<bool> dp(s.size()+1,false);
dp[0] = true;
for( int j = 0; j <= s.size(); j++){
//排列问题先遍历背包
for( int i = 0; i < j; i++){
string word = s.substr(i,j-i);
if( wordset.find(word) != wordset.end() && dp[i]){
dp[j] = true;
}
}
}
return dp[s.size()];
}
};
多重背包
多重背包可以直接转换为01背包,如下:
背包最大重量为10。
物品为:
重量 | 价值 | 数量 | |
---|---|---|---|
物品0 | 1 | 15 | 2 |
物品1 | 3 | 20 | 3 |
物品2 | 4 | 30 | 2 |
问背包能背的物品最大价值是多少?
和如下情况有区别么?
重量 | 价值 | 数量 | |
---|---|---|---|
物品0 | 1 | 15 | 1 |
物品0 | 1 | 15 | 1 |
物品1 | 3 | 20 | 1 |
物品1 | 3 | 20 | 1 |
物品1 | 3 | 20 | 1 |
物品2 | 4 | 30 | 1 |
物品2 | 4 | 30 | 1 |
没有任何区别,leetcode上找不到多重背包的题目
总结:将多重背包转换为01背包即可
// 时间复杂度:O(m*n*k),m为物品的种类,n背包容量,k单个物品数量
void test_multi_pack() {
vector<int> weight = {1, 3, 4};
vector<int> value = {15, 20, 30};
vector<int> nums = {2, 3, 2};
int bagWeight = 10;
for (int i = 0; i < nums.size(); i++) {
while (nums[i] > 1) { // nums[i]保留到1,把其他物品都展开
weight.push_back(weight[i]);
value.push_back(value[i]);
nums[i]--;
}
}
vector<int> dp(bagWeight + 1, 0);
for(int i = 0; i < weight.size(); i++) { // 遍历物品
for(int j = bagWeight; j >= weight[i]; j--) { // 遍历背包容量
dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
}
for (int j = 0; j <= bagWeight; j++) {
cout << dp[j] << " ";
}
cout << endl;
}
cout << dp[bagWeight] << endl;
}
int main() {
test_multi_pack();
}
另外一种方式
void test_multi_pack() {
vector<int> weight = {1, 3, 4};
vector<int> value = {15, 20, 30};
vector<int> nums = {2, 3, 2};
int bagWeight = 10;
vector<int> dp(bagWeight + 1, 0);
for(int i = 0; i < weight.size(); i++) { // 遍历物品
for(int j = bagWeight; j >= weight[i]; j--) { // 遍历背包容量
// 以上为01背包,然后加一个遍历个数
for (int k = 1; k <= nums[i] && (j - k * weight[i]) >= 0; k++) { // 遍历个数
dp[j] = max(dp[j], dp[j - k * weight[i]] + k * value[i]);
}
}
// 打印一下dp数组
for (int j = 0; j <= bagWeight; j++) {
cout << dp[j] << " ";
}
cout << endl;
}
cout << dp[bagWeight] << endl;
}
int main() {
test_multi_pack();
}
背包总结
动态规划五部曲:
1.确定dp[j]及下标的含义:这里一直用j就不改了
2.确定递推公式
3.dp数组初始化
4.确定遍历顺序
5.打印dp数组