[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