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 from left to right.
- The first integer of each row is greater than the last integer of the previous row.
For example,
Consider the following matrix:
[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return true.
先对矩阵的行数进行二分法查找， 找到target所在行数 再对此行二分法查找 来判断target时候存在。
这个的算法时间复杂度是O(log(rows)+log(columns))。
做这道题时候犯二了 在第一次二分法的时候先判断matrix[start][0] < target结果死活不对， 后来发现如果这么判定的话当matrix[end][0] < target 时候也会跳到start行。
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0){
return false;
}
if (matrix[0] == null || matrix[0].length == 0){
return false;
}
int start = 0;
int end = matrix.length - 1;
int row;
while (start + 1 < end){// find the row
int mid = start + (end - start) / 2;
if (matrix[mid][0] == target){
return true;
} else if (matrix[mid][0] < target){
start = mid;
} else {
end = mid;
}
}
if (matrix[end][0] <= target) {
row = end;
} else if (matrix[start][0] <= target) {
row = start;
} else {
return false;
}
start = 0;
end = matrix[0].length - 1;
while (start + 1 < end){// find the column
int mid = start + (end - start) / 2;
if (matrix[row][mid] == target){
return true;
} else if (matrix[row][mid] < target){
start = mid;
} else {
end = mid;
}
}
if (matrix[row][start] == target){
return true;
} else if (matrix[row][end] == target){
return true;
} else {
return false;
}
}
}
第二种方法
把2d矩阵转换成1d 赋值start = 0 end = row * column -1
每个元素都可以用 matrix[position/column][position%column]来表示
然后用2分法解题
时间复杂度 O(log(row * column))
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0){
return false;
}
if (matrix[0] == null || matrix[0].length == 0){
return false;
}
int row = matrix.length;
int column = matrix[0].length;
int start = 0;
int end = row * column - 1;
while (start + 1 < end){
int mid = start + (end - start) / 2;
int num = matrix[mid / column][mid % column];
if (num == target){
return true;
} else if(num < target){
start = mid;
} else {
end = mid;
}
}
if (matrix[start / column][start % column] == target){
return true;
} else if (matrix[end / column][end % column] == target){
return true;
} else {
return false;
}
}
}
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) {
return false;
}
if (matrix[0] == null || matrix[0].length == 0) {
return false;
}
int m = matrix.length;
int n = matrix[0].length;
int total = m * n;
int start = 0;
int end = total - 1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (matrix[mid/n][mid%n] == target) {
return true;
}
else if (matrix[mid/n][mid%n] > target) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return false;
}
}
