** problem **
(The length of each one-dimensional array of same), each row from left to right in order of ascending sort, to sort each column in a two-dimensional array in order of increasing from top to bottom. It determines whether it contains an array of integers.
** Examples **
** thinking **
- Violence O (n ^ 2)
The two-dimensional array of all the elements taken out into the set, and see whether there is the number set in
- Very violent
From top left [0] [0], and the number of elements taken comparison, if greater than, or let i + 1 j + 1, then ratio, is less than, let i-1 or j-1 ratio then
- Binary search O (nlogn)
Each line is increasing from left to right on each row binary search
- O (n)
- From the bottom left corner of view, up diminishing numbers, the right number can add up, so start looking from the lower left corner, when you are looking for a large digital numbers than the lower-left corner. To find the right figure than the lower left corner of the digital hours on shift.
- Compared with the method, each time the comparisons less a column or row, the comparison region greatly reduced
** ** Code
//方法3
public class Solution {
//二分查找，递归
public boolean binarySearch(int[] array, int low, int high,int target) {
//low>high要放在最前面，因为后面的[low],[high]可能会有越界
if (low>high || target<array[low] || target>array[high] ) return false;
int mid = (low+high)/2;
if (array[mid]>target) return binarySearch(array,low,mid-1,target);
if (array[mid]<target) return binarySearch(array,mid+1,high,target);
//if (array[mid]==target)
return true;
}
public boolean Find(int target, int [][] array) {
for (int i=0; i<array.length; i++) {
if (binarySearch(array[i],0,array[i].length-1,target)){
return true;
}
}
return false;
}
}
//方法4
import java.util.*;
public class Solution {
public boolean Find(int target, int [][] array) {
int i=array.length-1,j=0;
while (j<=array[0].length-1 && i>=0) {
int now= array[i][j];
if (now==target) return true;
if (now<target) j++;
if (now>target) i--;
/*
不用if(array[i][j]<target) j++;因为如果对i和j进行了修改，影响后面的if
否则的话，要写：
if(arr[i][j]<target) j++;
if (j<=array[0].length-1 && array[i][j]>target) i--;
*/
}
return false;
}
}
class Solution:
def searchMatrix(self, matrix, target):
"""
:type matrix: List[List[int]]
:type target: int
:rtype: bool
"""
if matrix==None or len(matrix)==0 or len(matrix[0])==0:
return False
i=len(matrix)-1;j=0;j_max=len(matrix[0])-1
while (i>=0 and j<=j_max):
t = matrix[i][j]
if target==t: return True
if target<t: i-=1
if target>t: j+=1
return False
** Tips **
Starting from the example, look at how you can reduce the number of operations to perform basic
