240 Search a 2D Matrix II

遍历二位数组

以下是题目：

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Given target = 5, return true.
Given target = 20, return false.

bool searchMatrix(int** matrix, int matrixRowSize, int matrixColSize, int target) { /*关于这道题目这是一种非常傻逼的解法,时间复杂度是O(m * n)*/     for (int i = 0; i < matrixRowSize; i++) {        for (int j = 0; j < matrixColSize; j++) {            if (matrix[i][j] == target)                return true;        }     }     return false; } bool searchMatrix(int** matrix, int matrixRowSize, int matrixColSize, int target) {    /*这是参考别人的优化之后的解法，这个解法的时间复杂度是O(m + n),充分了用了矩阵从左到右升序排列，从上到下升序排列的特性*/    int i = 0;    int j = matrixColSize - 1;    while (i < matrixRowSize && j >= 0) {        if (matrix[i][j] == target)            return true;        else if (matrix[i][j] > target)            j--;        else            i++;    }        return false; }