简直offer

数组中重复的数字

//O n O 1
//元素值与角标值 有明显对应关系
//维护一个指针
class Solution {
public:
    int duplicateInArray(vector<int>& nums) {
        int n = nums.size();
        for (auto x : nums)
            if (x < 0 || x >= n)
                return -1;

        for (int i = 0; i < nums.size(); i ++ ) {
            while (nums[i] != i) {
                if (nums[i] == nums[nums[i]])
                    return nums[i];
                swap(nums[i], nums[nums[i]]);
            }
        }

        return -1;
    }
};

二维数组中的查找

class Solution {
public:
    bool searchArray(vector<vector<int>>& matrix, int target) {
        if (matrix == NULL || matrix.empty() || matrix[0].empty()) return false;
        int i = 0, j = matrix[0].size() - 1;
        while (i < matrix.size() && j >= 0)
        {
            int t = matrix[i][j];
            if (t == target) return true;
            if (t > target) j -- ;
            else i ++ ;
        }
        return false;
    }
};

替换空格

public String replaceSpace(StringBuffer str) {
    int P1 = str.length() - 1;
    for (int i = 0; i <= P1; i++)
        if (str.charAt(i) == ‘ ‘)
            str.append("  ");

    int P2 = str.length() - 1;
    while (P1 >= 0 && P2 > P1) {
        char c = str.charAt(P1--);
        if (c == ‘ ‘) {
            str.setCharAt(P2--, ‘0‘);
            str.setCharAt(P2--, ‘2‘);
            str.setCharAt(P2--, ‘%‘);
        } else {
            str.setCharAt(P2--, c);
        }
    }
    return str.toString();
}

顺时针打印矩阵

public ArrayList<Integer> printMatrix(int[][] matrix) {
    ArrayList<Integer> ret = new ArrayList<>();
    int r1 = 0, r2 = matrix.length - 1, c1 = 0, c2 = matrix[0].length - 1;
    while (r1 <= r2 && c1 <= c2) {
        // 上
        for (int i = c1; i <= c2; i++)
            ret.add(matrix[r1][i]);
        // 右
        for (int i = r1 + 1; i <= r2; i++)
            ret.add(matrix[i][c2]);
        if (r1 != r2)
            // 下
            for (int i = c2 - 1; i >= c1; i--)
                ret.add(matrix[r2][i]);
        if (c1 != c2)
            // 左
            for (int i = r2 - 1; i > r1; i--)
                ret.add(matrix[i][c1]);
        r1++; r2--; c1++; c2--;
    }
    return ret;
}

重建二叉树

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
//前序遍历和中序遍历
class Solution {
public:
    unordered_map<int,int> pos;

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int n = preorder.size();
        for (int i = 0; i < n; i ++ )
            pos[inorder[i]] = i;  //定位根 在中序遍历中的位置
        return dfs(preorder, inorder, 0, n - 1, 0, n - 1);
    }

    TreeNode* dfs(vector<int>&pre, vector<int>&in, int pl, int pr, int il, int ir)
    {
        if (pl > pr) return NULL;
        int k = pos[pre[pl]] - il;
        TreeNode* root = new TreeNode(pre[pl]);
        root->left = dfs(pre, in, pl + 1, pl + k, il, il + k - 1);
        root->right = dfs(pre, in, pl + k + 1, pr, il + k + 1, ir);
        return root;
    }
};

二叉树的下一个节点 中序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode *father;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL), father(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode *p) {
        if (!p) return NULL;
        if (p->right) {
            p = p->right;
            while (p->left) p = p->left;
            return p;
        }

        while (p->father && p->father->right == p) {
            p = p->father;
        }

        if (p->father) return p->father;
        return NULL;
    }
};

矩阵中的路径

class Solution {
public:
    bool hasPath(vector<vector<char>>& matrix, string str) {
        for (int i = 0; i < matrix.size(); i ++ )
            for (int j = 0; j < matrix[i].size(); j ++ )
                if (dfs(matrix, str, 0, i, j))
                    return true;
        return false;
    }

    bool dfs(vector<vector<char>> &matrix, string &str, int u, int x, int y) {
        if (matrix[x][y] != str[u]) return false;
        if (u == str.size() - 1) return true;
        int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
        char t = matrix[x][y];
        matrix[x][y] = ‘*‘;    //每个点只走一次
        for (int i = 0; i < 4; i ++ ) {
            int a = x + dx[i], b = y + dy[i];
            if (a >= 0 && a < matrix.size() && b >= 0 && b < matrix[a].size()) {
                if (dfs(matrix, str, u + 1, a, b)) return true;
            }
        }
        matrix[x][y] = t;  //恢复
        return false;
    }
};

简直offer

原文：https://www.cnblogs.com/CSE-kun/p/14682913.html