题目描述:
解题思路:
方法1:遍历
public class Solution {
public boolean Find(int target, int [][] array) {
if(array.length<=0){
return false;
}
for(int i = array.length-1; i >= 0; i--){
for(int j = 0; j < array[0].length; j++){
if(array[i][j] == target){
return true;
}
}
}
return false;
}
}
方法2、
public class Solution {
public boolean Find(int target, int [][] array) {
if(array.length<=0){
return false;
}
int r = array.length-1;
int c = array[0].length-1;
int i = 0;
int j = c;
while(i <= c && j >= 0){
if(target < array[i][j]){
j--;
}else if(target > array[i][j]){
i++;
}else{
return true;
}
}
return false;
}
}
两个方法的时间复杂度