leetcode JAVA Set Matrix Zeroes 难度系数3 3.24

yundomo

于 2014-02-01 11:28:52 发布

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本文提供了一种高效的算法，在不使用额外空间的情况下，将矩阵中所有值为零的元素对应的行和列置零。通过仅利用矩阵的前两行和两列来标记需要置零的行和列，实现算法的空间复杂度为常数级。

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Question:

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.

click to show follow up.

Follow up:

Did you use extra space?
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?

public class Solution {
    public void setZeroes(int[][] matrix) {
		if (matrix.length < 1)
			return;
		boolean firstRow = false;
		boolean firstColumn = false;
		int row = matrix.length;
		int col = matrix[0].length;
		for (int i = 0; i < row; i++) {
			if (matrix[i][0] == 0) {
				firstColumn = true;
				break;
			}
		}
		for (int j = 0; j < col; j++) {
			if (matrix[0][j] == 0) {
				firstRow = true;
				break;
			}
		}
		for (int i = 1; i < row; i++) {
			for (int j = 1; j < col; j++) {
				if (matrix[i][j] == 0) {
					matrix[i][0] = 0;
					matrix[0][j] = 0;
				}
			}
		}
		for (int i = 1; i < row; i++) {
			for (int j = 1; j < col; j++) {
				if (matrix[i][0] == 0 || matrix[0][j] == 0) {
					matrix[i][j] = 0;
				}
			}
		}
		if (firstRow) {
			for (int j = 0; j < col; j++) {
				matrix[0][j] = 0;
			}
		}
		if (firstColumn) {
			for (int i = 0; i < row; i++) {
				matrix[i][0] = 0;
			}
		}
	}
}