[Leetcode]Search a 2D Matrix II

Search a 2D Matrix II My Submissions Question
Total Accepted: 19946 Total Submissions: 64757 Difficulty: Medium
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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当时看到这道题 我的想法是这样的，如下代码（p.s 实现的时候遇到了各种细节问题，下面是discuss中的）

class Solution {public:    bool search(vector<vector<int>> &matrix,int si,int ei,int sj,int ej,int target){        if(si>ei || sj>ej || matrix[si][sj] > target || matrix[ei][ej] < target)  return false;    //si is up side,ei is down side,sj is left side, ej is right side         int mi = (si + ei) / 2;        int mj = (sj + ej) / 2;        if(matrix[mi][mj] == target)            return true;        else if(matrix[mi][mj] < target){            return search(matrix,si,mi,mj+1,ej,target) ||search(matrix,mi+1,ei,sj,mj,target)||search(matrix,mi+1,ei,mj+1,ej,target);        }        else{            return search(matrix,si,mi-1,mj,ej,target) ||search(matrix,mi,ei,sj,mj-1,target) ||search(matrix,si,mi-1,sj,mj-1,target);        }    }    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if(matrix.empty()) return false;        return search(matrix,0,matrix.size()-1,0,matrix[0].size()-1,target);    }};

后来看到大神写的是这样的。。。

 bool searchMatrix(vector<vector<int>>& matrix, int target) {    int m = matrix.size(), n = matrix[0].size();    int i = 0, j = n - 1;    // search target from up-right side to down-left side    while (i <= m-1 && j >= 0) {        if (matrix[i][j] == target)            return true;        else if (matrix[i][j] > target)            --j;        else // matrix[i][j] < target            ++i;    }    return false;}

果然还停留在菜鸟水平，做完怀疑了一晚的智商