今日记录
62.不同路径
Leetcode链接
1.确定dp数组(dp table)以及下标的含义:dp[i][j] 含义 到达 (i,j) 处有 dp[i][j] 种路径
2.确定递推公式:dp[i][j] = dp[i-1][j] + dp[i][j-1]
3.dp数组如何初始化: for i=0 ~ m-1, dp[i][0]=1, for j=0 ~ n-1, dp[0][j]=1
4.确定遍历顺序:从左向右,从上往下
5.举例推导dp数组
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> dp(m, vector<int>(n, 0));
for (int i = 0; i < m; i++) {
dp[i][0] = 1;
}
for (int j = 0; j < n; j++) {
dp[0][j] = 1;
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
};
63.不同路径Ⅱ
1.确定dp数组(dp table)以及下标的含义: dp[i][j] 含义 到达 (i,j) 处有 dp[i][j] 种路径
2.确定递推公式: if (obstacleGrid[i][j] == 1 ) comtinue; dp[i][j] = dp[i-1][j] + dp[i][j-1]
3.dp数组如何初始化: for i=0 ~ m-1 且 obstacleGrid[i][0] == 0, dp[i][0]=1, for j=0 ~ n-1 且 obstacleGrid[0][j] == 0, dp[0][j]=1
4.确定遍历顺序
5.举例推导dp数组
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int>> dp(m, vector<int>(n, 0));
// 初始化
for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++)
dp[i][0] = 1;
for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++)
dp[0][j] = 1;
// 遍历循环
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1)
continue;
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
};