标签：遍历 width its leetcode sum round 计算 code each
Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.
Example 1:
Input: [[1,1,1], [1,0,1], [1,1,1]] Output: [[0, 0, 0], [0, 0, 0], [0, 0, 0]] Explanation: For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0 For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0 For the point (1,1): floor(8/9) = floor(0.88888889) = 0
Note:
对每一个点及其周围的值求平均值作为新的值赋予该点。首先要判断一个点周围有多少个有效点。利用isVaild判断有效点。然后遍历原二维数组，然后计算每个点周围的有效点平均值赋给该点后压入数组。
class Solution { public: vector<vector<int>> imageSmoother(vector<vector<int>>& M) { vector<vector<int>> res; int rows = M.size(), cols = M[0].size(); if (rows == 0 || cols == 0) return res; for (int i = 0; i != rows; i++) { vector<int> cur; for (int j = 0; j != cols; j++) { int sum = 0; int cnt = 0; for (int x = -1; x < 2; x++) { for (int y = -1; y < 2; y++) { if (isVaild(i + x, j + y, M)) { cnt++; sum += M[x + i][y + j]; } } } cur.push_back(sum / cnt); } res.push_back(cur); } return res; } bool isVaild(int i, int j, vector<vector<int>>& M) { if (i >= 0 && i < M.size() && j >= 0 && j < M[0].size()) return true; return false; } }; // 169 ms
标签：遍历 width its leetcode sum round 计算 code each
原文地址：http://www.cnblogs.com/immjc/p/7507869.html
