翻译

写一个高效算法用于在一个m x n的矩阵中查找一个值。这个矩阵有如下属性：每行的整型数都是从左到右排序的。每行的第一个元素都比上一行的最后一列大。例如，考虑如下矩阵：[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]给定target = 3，返回true。

原文

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.

分析

可能因为我好困了，所以不论是算法还是我自己，都效率很低……

下面这个代码也是一改再改……

bool searchMatrix(vector
     
      
       >& matrix, int target) {    if (matrix[0][0] > target) return false;    for (int i = 0; i < matrix.size(); ) {        for (int j = 0; j < matrix[i].size(); ) {            if (i == matrix.size() - 1 && matrix[i][j] < target) {                if (j >= matrix[i].size()) return false;                j += 1;                if (matrix[i][j] > target) return false;            }            else if (matrix[i][j] < target && matrix[i+1][j] > target) {                j += 1;                if (j >= matrix[i].size()) return false;                if (matrix[i][j] > target) return false;            }            else if (matrix[i][j] < target && matrix[i + 1][j] <= target) {                i += 1;            }            else if (matrix[i][j] == target) {                return true;            }                 if (i == matrix.size() - 1 && j == matrix[i].size() ) {                return false;            }        }    }    return false;}
      
     

明天再整理整理思路重新做一遍吧……