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数学、计算几何、位运算常见问题详解

分类: IT文章 • 2022-04-04 19:43:46

数学、计算几何、位运算常见问题详解

• 矩阵上的问题（3题）

Search a 2D Matrix II

    public int searchMatrix(int[][] matrix, int target) {
        // write your code here
        int n = matrix.length;
        if (n == 0) {
            return 0;
        }
        int m = matrix[0].length;
        if (m == 0) {
            return 0;
        }
        int i = n - 1;
        int j = 0;
        int res = 0;
        while (i >= 0 && j < m) {
            if (matrix[i][j] == target) {
                res++;
                i--;
                j++;
            } else if (matrix[i][j] > target) {
                i--;
            } else {
                j++;
            }
        }
        return res;
    }
View Code

Rotate Image

    public void rotate(int[][] matrix) {
        // write your code here
        int n = matrix.length;
        for (int i = 0; i < n / 2; i++) {
            for (int j = 0; j < n; j++) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[n - 1 - i][j];
                matrix[n - 1 - i][j] = temp;
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                int temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }
    }
View Code

Sparse Matrix Multiplication

     public int[][] multiply(int[][] a, int[][] b) {
        int n = a.length;
        int m = b[0].length;
        int t = a[0].length;
        int[][] res = new int[n][m];
        List<List<Integer>> col = new ArrayList<>();
        List<List<Integer>> val = new ArrayList<>();
        for (int i = 0; i < t; i++) {
            col.add(new ArrayList<>());
            val.add(new ArrayList<>());
            for (int j = 0; j < m; j++) {
                if (b[i][j] != 0) {
                    col.get(i).add(j);
                    val.get(i).add(b[i][j]);
                }
            }
        }
        for (int i = 0; i < n; i++) {
            for (int k = 0; k < t; k++) {
                if (a[i][k] == 0) {
                    continue;
                }
                for (int j = 0; j < col.get(k).size(); j++) {
                    int c = col.get(k).get(j);
                    int v = col.get(k).get(j);
                    res[i][j] += a[i][k] * v; 
                }
            }
        }
    }
View Code错 
public int[][] multiply(int[][] a, int[][] b){
        int m = a.length;
        int n = b[0].length;
        int t = a[0].length;
        List<List<Integer>> col = new ArrayList<>();
        List<List<Integer>> val = new ArrayList<>();
        for (int i = 0; i < t; i++) {
            col.add(new ArrayList<>());
            val.add(new ArrayList<>());
            for (int j = 0; j < n; j++) {
                if (b[i][j] != 0) {
                    col.get(i).add(j);
                    val.get(i).add(b[i][j]);
                }
            }
        }
        int[][] res = new int[m][n];
        for (int i = 0; i < m; i++) {
                for (int k = 0; k < t; k++) {
                    if (a[i][k] == 0) {
                        continue;
                    }
                    for (int p = 0; p < col.get(k).size(); p++) {
                        int j = col.get(k).get(p);
                        int v = val.get(k).get(p);
                        res[i][j] += a[i][k] * v;
                    }
                }
            }
        return res;
    }
View Code

• 高精度运算（4题）

Big Integer Addition 

    public String addStrings(String num1, String num2) {
        // write your code here
        String res = "";
        int carry = 0;
        for (int i = num1.length() - 1, j = num2.length() - 1; i >= 0 || j >= 0; i--, j--) {
            int sum = carry;
            sum += i >= 0 ? num1.charAt(i) - '0' : 0;
            sum += j >= 0 ? num2.charAt(j) - '0' : 0;
            res = sum % 10 + res;
            carry = sum / 10;
        }
        if (carry > 0) {
            res = carry + res;
        }
        return res;
    }
View Code

Big Integer multiplication

    public String multiply(String num1, String num2) {
        // write your code here
        int l1 = num1.length();
        int l2 = num2.length();
        int[] res = new int[l1 + l2 + 1];
        for (int i = 0; i < l1; i++) {
            for (int j = 0; j < l2; j++) {
                res[i + j] += (num1.charAt(l1 - 1 - i) - '0') * (num2.charAt(l2 - 1 - j) - '0');
            }
        }
        for (int i = 0; i < l1 + l2; i++) {
            res[i + 1] += res[i] / 10;
            res[i] = res[i] % 10;
        }
        int i = l1 + l2;
        while (res[i] == 0 && i >= 1) {
            i--;
        }
        String str = "";
        while (i >= 0) {
            str += res[i--];
        }
        return str;
    }
View Code

Add Binary

    public String addBinary(String a, String b) 
    {
        String res = "";
        int carry = 0;
        for (int i = a.length() - 1, j = b.length() - 1; i >= 0 || j >= 0; j--, i--) {
            int sum = carry;
            sum += i >= 0 ? (a.charAt(i) - '0') : 0;
            sum += j >= 0 ? (b.charAt(j) - '0') : 0;
            res = sum % 2 + res;
            carry = sum / 2;
        }
        if (carry > 0) {
            res = carry  + res;
        }
        return res;
    }
View Code

Add Two Numbers

    public ListNode addLists(ListNode l1, ListNode l2) 
    {
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        int carry = 0;
        for (ListNode i = l1, j = l2; i != null || j != null;) {
            int sum = carry;
            sum += i != null ? i.val : 0;
            sum += j != null ? j.val : 0;
            tail.next = new ListNode(sum % 10);
            tail = tail.next;
            i = i == null ? i : i.next;
            j = j == null ? j : j.next;
            carry = sum / 10;
        }
        if (carry > 0) {
            tail.next = new ListNode(carry);
        }
        return dummy.next;
    }
View Code

• 快速幂（1题）

Pow(x, n)

     public double myPow(double x, int n) 
    {
        if (n < 0) {
            x = 1 / x;
            n = -n;
        }
        double res = 1, temp = x;
        while (n != 0) {
            if (n % 2 == 1) {
                res *= temp;
            }
            n /= 2;
            temp = temp * temp;
        }
        return res;
    }
View Code 
     public double myPow(double x, int m) 
    {
        long n = m;
        if (n < 0) {
            x = 1 / x;
            n = -n;
        }
        double res = 1, temp = x;
        while (n != 0) {
            if (n % 2 == 1) {
                res *= temp;
            }
            n /= 2;
            temp = temp * temp;
        }
        return res;
    }
View Code

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