[LeetCode 74]Search a 2D Matrix

题目链接：search-a-2d-matrix

/** * 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.For example,Consider the following matrix:[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]Given target = 3, return true. * */public class SearchA2DMatrix {//134 / 134 test cases passed.//Status: Accepted//Runtime: 236 ms//Submitted: 0 minutes ago//时间复杂度 O(log(max(m,n)) 空间复杂度 O(1)    static boolean searchMatrix(int[][] matrix, int target) {               int m = matrix.length; //行数        int n = matrix[0].length;//列数        int up = 0;        int down = m - 1;        int left = 0;        int right = n;                //二分法查找所在行        while(up < down) {        int mid = (down + up) / 2;        if(matrix[mid][n - 1] > target) down = mid;        else if(matrix[mid][n - 1] < target) up = mid + 1;        else return true;        }        //二分法查找所在列        while(left < right) {        int mid = (left + right) / 2;        if(matrix[up][mid] > target) right = mid;        else if(matrix[up][mid] < target) left = mid + 1;        else return true;        }            return false;    }public static void main(String[] args) {System.out.println(searchMatrix(new int[][]{{1, 3,  5,  7},{10, 11, 16, 20},{23, 30, 34, 50}}, 3));System.out.println(searchMatrix(new int[][]{{1, 1}}, 2));System.out.println(searchMatrix(new int[][]{{1,}, { 3}}, 2));System.out.println(searchMatrix(new int[][]{{1}}, 1));}}