Sword finger offer (II) -- search in two-dimensional array

Title Description

In a two-dimensional array （ Each one-dimensional array has the same length ）, Each row is sorted in ascending order from left to right , Each column is sorted in ascending order from top to bottom . Please complete a function , Enter such a two-dimensional array and an integer , Determine whether the array contains the integer .

Example

Enter an array ：

num[3][4]=[ 1,4,6,28, 2,7,32,30, 10,11,67,79 ]

You need to find a number 32, Then return to true

Ideas

You can go through it directly , But this complexity in the worst case is to facilitate all to get the results . If the array size is n×m, So the complexity is O(n×m). But let's think another way , We chose the one in the lower left corner 10(num[2][0],i=2,j=0) As a starting point , If it is greater than 10, that i+1, If it is less than 10, be j+1, The next number to look up is 11, We know 32 Still more than 11 Big , To the right 67, then 32 Than 67 Small , We should look up , eureka 32. If you look for 28, It's the worst , Find the end of the upper right corner of the array , thus , The worst result is O（n+m）.

Code

public class Solution {    public boolean Find(int target, int [][] array) {            int size=array.length;            int length=array[0].length;            int i,j;            for(i=size-1,j=0;i>=0&&i<size&&j>=0&&j<length;){                if(array[i][j]==target){                    return true;
}                if(array[i][j]<target){
j++;                    continue;
}                if(array[i][j]>target){
i--;                    continue;
}
}        return false;
}
}