322. 零钱兑换
class Solution {
public:int coinChange(vector<int>& coins, int amount) {//min countvector<int> dp(amount+1, INT_MAX);int n = coins.size();//initdp[0] = 0;//不包含顺序, 先物品再容量for (int i = 0; i < n; i++) {for (int j = coins[i]; j <= amount; j++) {if (dp[j - coins[i]] != INT_MAX)dp[j] = min(dp[j], dp[j - coins[i]]+1);}}if (dp[amount] == INT_MAX) return -1;return dp[amount];}};
279. 完全平方数
class Solution {
public:int numSquares(int n) {//dp 和为i的完全平方数的最少数量vector<int> dp(n+1, INT_MAX);dp[0] = 0;//没有顺序，先物品再容量for (int i = 1; i*i <= n; i++) {for (int j = i*i; j <= n; j++ ) {dp[j] = min(dp[j - i*i]+1, dp[j]);}}return dp[n];}
};
139. 单词拆分
class Solution {
public:bool wordBreak(string s, vector<string>& wordDict) {unordered_set<string> words(wordDict.begin(), wordDict.end());vector<bool> dp(s.length()+1, false);dp[0] = true;//有顺序，先容量再物品for (int i = 0; i <= s.size(); i++) {for (int j = 0; j < i; j++) {//取字串string str = s.substr(j, i - j);//若存在, 则返回trueif (words.find(str) != words.end() && dp[j]) {dp[i] = true;}}}return dp[s.size()];}
};
)
