leetcode打卡#day46 322. 零钱兑换、279. 完全平方数、139. 单词拆分

322. 零钱兑换

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        //min count
        vector<int> dp(amount+1, INT_MAX);
        int n = coins.size();
        //init
        dp[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true
                if (words.find(str) != words.end() && dp[j]) {
                    dp[i] = true;
                }
            }
        }
        return dp[s.size()];
    }
};