LeetCode 74. Search a 2D Matrix--python,java,c++解法

LeetCode 74. Search a 2D Matrix

---

LeetCode题解专栏：LeetCode题解
我做的所有的LeetCode的题目都放在这个专栏里，大部分题目Java和Python的解法都有。

---

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.
The first integer of each row is greater than the last integer of the previous row.
Example 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

---

官方文章;Search in 2D Matrix - LeetCode Articles
这道题目可以二分搜索查找：
python解法：

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        m = len(matrix)
        if m == 0:
            return False
        n = len(matrix[0])
        
        # binary search
        left, right = 0, m * n - 1
        while left <= right:
                pivot_idx = (left + right) // 2
                pivot_element = matrix[pivot_idx // n][pivot_idx % n]
                if target == pivot_element:
                    return True
                else:
                    if target < pivot_element:
                        right = pivot_idx - 1
                    else:
                        left = pivot_idx + 1
        return False

Java解法：

class Solution {
  public boolean searchMatrix(int[][] matrix, int target) {
    int m = matrix.length;
    if (m == 0) return false;
    int n = matrix[0].length;
    // binary search
    int left = 0, right = m * n - 1;
    int pivotIdx, pivotElement;
    while (left <= right) {
      pivotIdx = (left + right) / 2;
      pivotElement = matrix[pivotIdx / n][pivotIdx % n];
      if (target == pivotElement) return true;
      else {
        if (target < pivotElement) right = pivotIdx - 1;
        else left = pivotIdx + 1;
      }
    }
    return false;
  }
}

c++解法：

class Solution {
  public:
  bool searchMatrix(vector<vector<int>>& matrix, int target) {
    int m = matrix.size();
    if (m == 0) return false;
    int n = matrix[0].size();

    // binary search
    int left = 0, right = m * n - 1;
    int pivotIdx, pivotElement;
    while (left <= right) {
      pivotIdx = (left + right) / 2;
      pivotElement = matrix[pivotIdx / n][pivotIdx % n];
      if (target == pivotElement) return true;
      else {
        if (target < pivotElement) right = pivotIdx - 1;
        else left = pivotIdx + 1;
      }
    }
    return false;
  }
};