LeetCode算法题目：Search a 2D Matrix II

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.

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分析：

此题与Search 2D Matrix基本相同，但现在下一列的最小值不再大于上一列的最大值，故可以采用从左下角或者右上角开始搜索的方法。

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代码：

//从左下角开始查找，也可以从右上角开始查找class Solution {  public:      bool searchMatrix(vector<vector<int>>& matrix, int target) {          if (matrix.empty()) return false;        int n = matrix.size();          int m = matrix.front().size();          int i = n-1;          int j = 0;          while(i>=0 && j<m){              if (matrix[i][j] == target)                  return true;              else if (matrix[i][j]<target)                  j++;              else                  i--;          }          return false;      }  };  

//从右上角开始查找class Solution {  public:      bool searchMatrix(vector<vector<int>>& matrix, int target) {          if(matrix.empty()) return false;        int m = matrix.size();        int n = matrix.front().size();          int i=0, j=n-1;          while (i<m && j>=0)          {              int x = matrix[i][j];              if (x == target)                  return true;            else if (x >target)                  j--;              else                i++;          }          return false;      }  };