力扣（2）

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74. 搜索二维矩阵

思路：

代码：

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& mat, int t) {
        int m = mat.size(), n = mat[0].size();

        // 第一次二分：定位到所在行（从上往下，找到最后一个满足 mat[x]][0] <= t 的行号）
        int l = 0, r = m - 1;
        while (l < r) {
            int mid = (l + r + 1 )>> 1;
            if (mat[mid][0] <= t) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }

        int row = r;
        if (mat[row][0] == t) return true;
        if (mat[row][0] > t) return false;

        // 第二次二分：从所在行中定位到列（从左到右，找到最后一个满足 mat[row][x] <= t 的列号）
        l = 0; r = n - 1;
        while (l < r) {
            int mid = (l + r + 1 )>> 1;
            if (mat[row][mid] <= t) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        int col = r;

        return mat[row][col] == t;

    }
};

208. 实现 Trie (前缀树)

思路：

代码：

class Trie {
private:
    bool isEnd;
    Trie*next[26];
public:
    Trie() {
        isEnd = false;
        memset(next, 0, sizeof(next));
    }
    
    void insert(string word) {
        Trie*node = this;
        for (char c : word) {
            if (node->next[c-'a'] == NULL) {
                node->next[c-'a'] = new Trie();
            }
            node = node->next[c-'a'];
        }
        node->isEnd = true;
    }
    
    bool search(string word) {
        Trie*node = this;
        for (char c : word) {
            node = node->next[c-'a'];
            if (node == NULL) {
                return false;
            }
        }
        return node->isEnd;
    }
    
    bool startsWith(string prefix) {
        Trie*node = this;
        for (char c : prefix) {
            node = node->next[c-'a'];
            if (node == NULL) {
                return false;
            }
        }return true;
    }
};

/**
 * Your Trie object will be instantiated and called as such:
 * Trie* obj = new Trie();
 * obj->insert(word);
 * bool param_2 = obj->search(word);
 * bool param_3 = obj->startsWith(prefix);
 */

518. 零钱兑换 II

思路：

代码：

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        for (int i = 0; i < coins.size(); i++) {
            for (int j = coins[i]; j <= amount; j++) {
                dp[j] += dp[j - coins[i]];
            }
        }return dp[amount];
    }
};

剑指 Offer II 102. 加减的目标值

思路：

代码：

class Solution {
public:
    int findTargetSumWays(vector<int>& nums, int s) {
        int sum = 0;
        for (int i = 0; i < nums.size(); i++) {
            sum += nums[i];
        }
        if (abs(s) > sum) return 0;
        if ((s + sum) % 2 == 1) return 0;
        int bagSize = (s + sum) / 2;
        vector<int> dp(bagSize + 1, 0);
        dp[0] = 1;
        for (int i = 0; i <nums.size(); i++) {
            for (int j = bagSize; j >= nums[i]; j--) {
                dp[j] += dp[j - nums[i]];
            }
        }return dp[bagSize];
    }
};