24
07/2015
[LeetCode] Search a 2D Matrix II
Search a 2D Matrix II
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.
解题思路：
咋一看，还不知如何下手，因为这样不能像此前Search a 2D Matrix那样，将二维坐标转化成一维坐标。其实可以从二维的角度来将矩阵划分。
矩阵中心为9，其中左上角的所有元素均小于等于9，右下角的元素均大于9。因此，我们可以每次对比矩阵中心元素，然后排出左上角或右下角区域即可。下面是代码：
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m = matrix.size();
if(m<=0){
return false;
}
int n = matrix[0].size();
if(n<=0){
return false;
}
return searchMatrixHelper(matrix, 0, m-1, 0, n-1, target);
}
bool searchMatrixHelper(vector<vector<int>>& matrix, int startM, int endM, int startN, int endN, int target){
if(startM > endM || startN > endN){
return false;
}
int middleM = (startM + endM) / 2;
int middleN = (startN + endN) / 2;
if(matrix[middleM][middleN]==target){
return true;
}else if(matrix[middleM][middleN]<target){
return searchMatrixHelper(matrix, startM, endM, middleN + 1, endN, target) ||
searchMatrixHelper(matrix, middleM + 1, endM, startN, middleN, target);
}else{
return searchMatrixHelper(matrix, startM, middleM - 1, startN, endN, target) ||
searchMatrixHelper(matrix, middleM, endM, startN, middleN - 1, target);
}
}
};转载请注明：康瑞部落 » [LeetCode] Search a 2D Matrix II
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