Problem
+Given an m x n binary matrix mat, return **the distance of the nearest *0* for each cell**.
The distance between two adjacent cells is 1.
+Example 1:
+
Input: mat = [[0,0,0],[0,1,0],[0,0,0]]
+Output: [[0,0,0],[0,1,0],[0,0,0]]
+Example 2:
+
Input: mat = [[0,0,0],[0,1,0],[1,1,1]]
+Output: [[0,0,0],[0,1,0],[1,2,1]]
++Constraints:
+-
+
m == mat.length
+n == mat[i].length
+1 <= m, n <= 104
+1 <= m * n <= 104
+mat[i][j]is either0or1.
+There is at least one
0inmat.
+
Solution
+/**
+ * @param {number[][]} mat
+ * @return {number[][]}
+ */
+var updateMatrix = function(mat) {
+ var arr = [];
+ var m = mat.length;
+ var n = mat[0].length;
+ var res = Array(m).fill(0).map(() => Array(n));
+ var mark = function(i, j, distance) {
+ if (mat[i] === undefined || mat[i][j] === undefined) return;
+ if (res[i][j] !== undefined) return;
+ arr.push([i, j]);
+ res[i][j] = distance;
+ };
+ for (var i = 0; i < m; i++) {
+ for (var j = 0; j < n; j++) {
+ mat[i][j] === 0 && mark(i, j, 0);
+ }
+ }
+ while (arr.length) {
+ var [i, j] = arr.shift();
+ mark(i - 1, j, res[i][j] + 1);
+ mark(i + 1, j, res[i][j] + 1);
+ mark(i, j - 1, res[i][j] + 1);
+ mark(i, j + 1, res[i][j] + 1);
+ }
+ return res;
+};
+Explain:
+Breadth-first search.
+Complexity:
+-
+
- Time complexity : O(n). +
- Space complexity : O(n). +
+