常规做法dfs bfs O(mn)
bfs需要新建一个class存坐标 dfs不用存 更简便一点 记得每次访问过1后要给他设置成0
public class Solution { int minX = Integer.MAX_VALUE, maxX = 0, minY = Integer.MAX_VALUE, maxY = 0; public int minArea(char[][] image, int x, int y) { if ( image == null || image.length == 0 || image[ 0 ].length == 0 ) return 0; dfs ( image, x, y ); return ( maxX - minX + 1 ) * ( maxY - minY + 1 ); } public void dfs ( char[][] image, int x, int y ){ int m = image.length; int n = image[ 0 ].length; if ( x < 0 || x >= m || y < 0 || y >= n || image [x][y] == '0' ) return; image[x][y] = '0'; minX = Math.min( minX, x ); maxX = Math.max( maxX, x ); minY = Math.min( minY, y ); maxY = Math.max( maxY, y ); dfs ( image, x - 1, y ); dfs ( image, x + 1, y ); dfs ( image, x, y - 1 ); dfs ( image, x, y + 1 ); } }
还有更快的就是binary search 做四次 寻找边界
[0,y) 找left 找的是最左边的1
[y + 1, n) 找right 找的是右边第一个0
top bottom同理
那个boolean只是为了区分这次是往哪边找 不同方向start和end移动方式不一样
最后算面积的时候right - left不用再加1了
public class Solution { public int minArea(char[][] image, int x, int y) { int m = image.length; int n = image[ 0 ].length; int left = searchCol ( image, 0, y, true ); int right = searchCol ( image, y + 1, n, false ); int top = searchRow ( image, 0, x, true ); int bottom = searchRow ( image, x + 1, m, false ); return ( right - left ) * ( bottom - top ); } public int searchCol ( char[][]image, int start, int end, boolean left ){ int row = image.length; while ( start < end ){ int mid = start + ( end - start ) / 2; int k = 0; while ( k < row && image[ k ][ mid ] == '0') k ++; if ( k < row == left ) end = mid; else start = mid + 1; } return start; } public int searchRow ( char[][] image, int start, int end, boolean top ){ int col = image[ 0 ].length; while ( start < end ){ int mid = start + ( end - start )/ 2; int k = 0; while ( k < col && image [ mid ][ k ] == '0' ) k ++; if ( k < col == top ) end = mid; else start = mid + 1; } return start; } }