finish 73

Author: BaffinLee

73. Set Matrix Zeroes.js

@@ -0,0 +1,86 @@
+/**
+ * 73. Set Matrix Zeroes
+ * 
+ * Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.
+ * 
+ * Example 1:
+ * 
+ * Input: 
+ * [
+ *   [1,1,1],
+ *   [1,0,1],
+ *   [1,1,1]
+ * ]
+ * Output: 
+ * [
+ *   [1,0,1],
+ *   [0,0,0],
+ *   [1,0,1]
+ * ]
+ * Example 2:
+ * 
+ * Input: 
+ * [
+ *   [0,1,2,0],
+ *   [3,4,5,2],
+ *   [1,3,1,5]
+ * ]
+ * Output: 
+ * [
+ *   [0,0,0,0],
+ *   [0,4,5,0],
+ *   [0,3,1,0]
+ * ]
+ * Follow up:
+ * 
+ * A straight forward solution using O(mn) space is probably a bad idea.
+ * A simple improvement uses O(m + n) space, but still not the best solution.
+ * Could you devise a constant space solution?
+ */
+
+/**
+ * @param {number[][]} matrix
+ * @return {void} Do not return anything, modify matrix in-place instead.
+ */
+var setZeroes = function(matrix) {
+	var m = matrix.length;
+	var n = (matrix[0] || []).length;
+	for (var i = 0; i < m; i++) {
+			for (var j = 0; j < n; j++) {
+					if (matrix[i][j] === 0) {
+							left(i, j, m, n, matrix);
+							right(i, j, m, n, matrix);
+							up(i, j, m, n, matrix);
+							down(i, j, m, n, matrix);
+					} else if (matrix[i][j] === '#') {
+							matrix[i][j] = 0;
+					}
+			}
+	}
+};
+
+var left = function (i, j, m, n, matrix) {
+	for (var k = j - 1; k >= 0; k--) {
+			matrix[i][k] = 0;
+	}
+};
+
+var right = function (i, j, m, n, matrix) {
+	for (var k = j + 1; k < n; k++) {
+			matrix[i][k] = matrix[i][k] === 0 ? 0 : '#';
+	}
+};
+
+var up = function (i, j, m, n, matrix) {
+	for (var k = i - 1; k >= 0; k--) {
+			matrix[k][j] = 0;
+	}
+};
+
+var down = function (i, j, m, n, matrix) {
+	for (var k = i + 1; k < m; k++) {
+			matrix[k][j] = matrix[k][j] === 0 ? 0 : '#';
+	}
+};
+
+// 把没遍历的 1 设置为 0 会影响之后的判断，先设置为 # ，再改回来