Difficulty:: Medium
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.
Example 1:
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| Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true
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Example 2:
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| Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false
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Solution
Language: Java
两次二分法
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| class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return false;
}
int m = matrix.length;
int n = matrix[0].length;
int start = 0, end = m - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (matrix[mid][0] == target) {
return true;
} else if (matrix[mid][0] > target) {
end = mid;
} else {
start = mid;
}
}
int row = 0;
if (matrix[end][0] <= target) {
row = end;
} else {
row = start;
}
start = 0;
end = n - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (matrix[row][mid] == target) {
return true;
} else if (matrix[row][mid] > target) {
end = mid;
} else {
start = mid;
}
}
return matrix[row][start] == target || matrix[row][end] == target;
}
}
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不使用二分
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| class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0)
return false;
int r = 0;
int c = matrix[0].length-1;
while (c >= 0 && r < matrix.length) {
if (matrix[r][c] > target) {
c--;
} else if (matrix[r][c] < target) {
r++;
} else {
return true;
}
}
return false;
}
}
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