代码随想录第36天|动态规划

62. 不同路径

补充: 对二维数组的操作

dp[j][i] 表示到 j,i 有多少种路径
递推公式: dp[j][i] = dp[j - 1][i] + dp[j][i - 1]
初始化: dp[0][i] 和 dp[j][0] 都只有1种情况
遍历顺序: 由于dp[j][i] 由上和左的元素推导, 所以采用从左到右、从上到下的遍历顺序

class Solution {
public:void uniquePaths(int m, int n) {vector<vector<int>> dp(m, vector<int>(n, 0));//初始化for (int i = 0; i < dp[0].size(); i++) dp[0][i] = 1;for (int i = 0; i < dp.size(); i++) dp[i][0] = 1;//递推公式for (int j = 1; j < m; j++) {for (int i = 1; i < n; i++) {dp[j][i] = dp[j][i - 1] + dp[j - 1][i];}}//输出dp[j][i]for (int j = 0; j < dp.size(); j++) {for (int i = 0; i < dp[i].size(); i++) {cout << dp[j][i] << " ";}cout << endl;}return dp[m - 1][n - 1];                  }
};

63. 不同路径 II

参考

dp[j][i] 表示到(j, i) 后的路径数
递推公式
初始化 dp[0][i] 和 dp[j][0] 为 1 ,当遇到障碍物后变 0
遍历顺序同上

class Solution {
public:int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int n = obstacleGrid[0].size();int m = obstacleGrid.size();        vector<vector<int>> dp = vector<vector<int>>(m, vector<int>(n, 0));for (int i = 0; i < n; i++) {if (obstacleGrid[0][i] == 1) break;dp[0][i] = 1;}for (int j = 0; j < m; j++) {if (obstacleGrid[j][0] == 1) break;dp[j][0] = 1;}for (int j = 1; j < m; j++) {for (int i = 1; i < n; i++) {if (obstacleGrid[j][i] == 1) {//该点有障碍dp[j][i] = 0;                    } else {dp[j][i] = dp[j][i - 1] + dp[j - 1][i];}              }}return dp[m - 1][n - 1];}
};