标签：image pat http amp images grid mat order ret
1 63. Unique Paths II 带障碍物的路径计算
思路：dp[i][j] = 0 if grid[i][j] = 1 （障碍物）
再按照无障碍物的逻辑进行计算
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int rows = obstacleGrid.size(); if (rows == 0) { return 0; } int cols = obstacleGrid[0].size(); vector<vector<int>> ways(rows, vector<int> (cols, 1)); for (int i = 0; i < rows; i++) { for (int j = 0; j < cols; j++) { if (obstacleGrid[i][j] == 1) { ways[i][j] = 0; } else { if (i == 0 && j == 0) { ways[i][j] = 1; } else if (i == 0) { ways[i][j] = ways[i][j - 1]; } else if (j == 0) { ways[i][j] = ways[i - 1][j]; } else { ways[i][j] = ways[i - 1][j] + ways[i][j - 1]; } } } } return ways[rows - 1][cols - 1]; }
2 矩阵的环形遍历
方法：(1)设置rowStart,rowEnd,colStart,colEnd四个变量来指示当前轮遍历的起始位置
(2) 按照顺时针方向进行遍历，一轮遍历分成四次循环，每一次循环完成，终点或起点需要缩进一个单位(start + 1,end -1)
注意：当矩阵不是方阵时，需要在每一轮向左和向上遍历之前进行一次判断，否则会造成重复。
vector<int> spiralOrder(vector<vector<int>>& matrix) { if (matrix.size() == 0) { return {}; } int rowStart = 0; int rowEnd = matrix.size() - 1; int colStart = 0; int colEnd = matrix[0].size() - 1; vector<int> order; while (rowStart <= rowEnd && colStart <= colEnd) { for (int j = colStart; j <= colEnd; j++) { order.push_back(matrix[rowStart][j]); } rowStart++; for (int i = rowStart; i <= rowEnd; i++) { order.push_back(matrix[i][colEnd]); } colEnd--; if (rowStart <= rowEnd) { //注意此处 for (int j = colEnd; j >= colStart; j--) { order.push_back(matrix[rowEnd][j]); } rowEnd--; } if (colStart <= colEnd) {//注意此处 for (int i = rowEnd; i >= rowStart; i--) { order.push_back(matrix[i][colStart]); } colStart++; } } return order; }
标签：image pat http amp images grid mat order ret
原文地址：http://www.cnblogs.com/lrlthinking/p/6134800.html
