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1269. Number of plans that are stuck in place
Solution: Coordinate dynamic programming
class Solution {
public:
int numWays(int steps, int arrLen) {
if (arrLen <= 0) {
return 0;
}
// 因为需要返回到0下标位置所以，最远也就是一半
int len = std::min(steps/2+1, arrLen);
// 走了多少步，到达当前下标dp[step][j]，有多少种
std::vector<std::vector<int>> dp(steps+1, std::vector<int>(len, 0));
// 从上一步得来，
std::vector<std::vector<int>> direction{
{-1, 0}, {-1, -1}, {-1, 1}};
dp[0][0] = 1;
int MOD = 1000000007;
for (int i = 1; i <= steps; ++i) {
for (int j = 0; j < len; ++j) {
for (auto& direc : direction) {
int prev_row = i + direc[0];
int prev_col = j + direc[1];
if (prev_row < 0 || prev_col < 0 || prev_col >= len) {
continue;
}
dp[i][j] = (dp[i][j] + dp[prev_row][prev_col]) % MOD;
}
}
}
return dp[steps][0];
}
};class Solution {
public:
int numWays(int steps, int arrLen) {
if (arrLen <= 1) {
return 1;
}
int MOD = 1000000007;
int len = std::min(steps/2+1, arrLen);
std::vector<std::vector<int>> dp(steps+1, std::vector<int>(len, 0));
dp[0][0] = 1;
for (int step = 1; step <= steps; ++step) {
dp[step][0] = (dp[step-1][0] + dp[step-1][1])%MOD;
dp[step][len-1] = (dp[step-1][len-1] + dp[step-1][len-2])%MOD;
for (int i = 1; i < len-1; ++i) {
dp[step][i] = ((dp[step-1][i-1] + dp[step-1][i])%MOD + dp[step-1][i+1])%MOD;
}
}
return dp[steps][0];
}
};