302. Smallest Rectangle Enclosing Black Pixels
An image is represented by a binary matrix with 0 as a white pixel and 1 as a black pixel. The black pixels are connected, i.e., there is only one black region. Pixels are connected horizontally and vertically. Given the location (x, y) of one of the black pixels, return the area of the smallest (axis-aligned) rectangle that encloses all black pixels.
Example:
Input:
[
“0010”,
“0110”,
“0100”
]
and x = 0, y = 2
Output: 6
solution : binary search
class Solution {
public int minArea(char[][] image, int x, int y) {
int m = image.length;
int n = image[0].length;
int l = 0;
int r = 0;
int t = 0;
int b = 0;
// vertical (column)
// left
int lo = 0;
int hi = y;
while (lo < hi) {
int mid = lo + (hi - lo) / 2;
if (emptyColumn(image, mid)) {
lo = mid+1;
} else {
hi = mid;
}
}
l = lo;
lo = y;
hi = n - 1;
while (lo < hi) {
int mid = lo + (hi - lo + 1) / 2;
if (emptyColumn(image, mid)) {
hi = mid - 1;
} else {
lo = mid;
}
}
r = lo;
// horizontal (row)
lo = 0;
hi = x;
while (lo + 1 < hi) {
int mid = lo + (hi - lo) / 2;
if (emptyRow(image, mid)) {
lo = mid;
} else {
hi = mid;
}
}
t = (emptyRow(image, lo)) ? hi:lo;
lo = x;
hi = m - 1;
while (lo + 1 < hi) {
int mid = lo + (hi - lo) / 2;
if (emptyRow(image, mid)) {
hi = mid;
} else {
lo = mid;
}
}
b = (emptyRow(image, hi)) ? lo:hi;
return (b - t + 1) * (r - l + 1);
}
public boolean emptyRow(char[][] image, int num) {
for (char tmp : image[num]) {
if (tmp == '1')
return false;
}
return true;
}
public boolean emptyColumn(char[][] image, int num) {
int m = image.length;
for (int i = 0; i < m; i++) {
if (image[i][num] == '1')
return false;
}
return true;
}
}
最小包围矩形算法

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