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.
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.
题目大意：
在二维数组中找寻指定的数字是否存在，数组从左到右递增，数组从上到下递增。
解法：
我想的是一行一行的寻找，如果说行开头的数字大于目标数字，那么说明目标数在数组中不存在，返回false，如果行结尾的数字小于目标数，说明该行没有目标数字，到下一行去寻找，否则就对该行进行二分查找。
class Solution {
private boolean find(int[][] matrix,int i,int target){
int left=0,right=matrix[0].length-1;
while(left<=right){
int mid=(left+right)/2;
if (matrix[i][mid]==target) return true;
else if (matrix[i][mid]<target) left=mid+1;
else right=mid-1;
}
return false;
}
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length==0 || matrix[0].length==0) return false;
int m=matrix.length,n=matrix[0].length;
if (target<matrix[0][0]||target>matrix[m-1][n-1]) return false;
for(int i=0;i<m;i++){
if (matrix[i][0]>target) return false;
if (matrix[i][n-1]<target) continue;
if (find(matrix,i,target)){
return true;
}
}
return false;
}
}
从数组的右上角开始寻找，这么简单的题目，我竟然给忘记了。
class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix==null||matrix.length==0||matrix[0].length==0) return false;
int row=0,col=matrix[0].length-1;
while (row<matrix.length && col>=0){
if (matrix[row][col]==target) return true;
else if (matrix[row][col]<target) row++;
else col--;
}
return false;
}
}