【java】74. 搜索二维矩阵---代码优化，时间复杂度接近O(N)!!!
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编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性：
每行中的整数从左到右按升序排列。
每行的第一个整数大于前一行的最后一个整数。示例 1：
输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
输出：false提示：
m == matrix.length
n == matrix[i].length 1 <= m, n <= 100 -104 <= matrix[i][j], target <= 104代码：public boolean searchMatrix(int[][] matrix, int target) { for(int i=0;i =target) { for(int j=0;j 转载地址：http://tamq.baihongyu.com/
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