[LeetCode] Search a 2D Matrix
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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
从左上角到右下角的顺序遍历数组,如果左上角元素大于target,则左移一位,直到元素小于target时,开始从该列从上到下遍历,直到找到target。
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { if (matrix.empty()) return false; int m = matrix.size(); int n = matrix[0].size(); int i = 0, j = n-1; while (i < m && j >= 0) { if (matrix[i][j] == target) return true; else if (matrix[i][j] > target) --j; else if (matrix[i][j] < target) ++i; } return false; } };
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