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算法笔记之DP实战（二）
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最大最小值DP
Choose minimum (maximum) path among all possible paths before the current state, then add value for the current state.
routes[i] = min(routes[i-1], routes[i-2], ... , routes[i-k]) + cost[i] 有时候也会min(f[x-1]+1, f[x])， 特别时重复走过x
746. Min Cost Climbing Stairs
到i的路径来自于i-1和i-2。最后一步可以为n-1或者n的总cost。
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
int ss = cost.size();
vector <int> memo (ss+1, 0x6fffffff);
memo[0] = 0; memo[1] = cost[0];
for (int i=2; i<=ss; i++) {
memo[i] = min(memo[i-1], memo[i-2]) + cost[i-1];
}
return min(memo[ss-1], memo[ss]);
}
};
64. Minimum Path Sum
Note: You can only move either down or right at any point in time.
f[n][m] = min(f[n-1][m], f[n][m-1]) + f[n][m];
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
vector<vector<int>> memo(n+1, vector<int>(m+1));
// 給memo[0][1], memo[1][0]赋初值0
for (int i=2; i<=n; i++) {
memo[i][0] = 0x6fffffff;
}
for (int i=2; i<=m; i++) {
memo[0][i] = 0x6fffffff;
}
for (int i=1; i<=n; i++) {
for (int j=0; j<=m; j++) {
memo[i][j] = min(memo[i-1][j], memo[i][j-1]) + grid[i-1][j-1];
}
}
return memo[n][m];
}
};
改良版->不需要memo, 用if处理padding，
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int n = grid.size(), m = grid[0].size();
// f[n][m] = min(f[n-1][m], f[n][m-1]) + f[n][m];
for (int i=0; i<n; i++) {
for (int j=0; j<m; j++) {
if (j>0 && i>0)
grid[i][j] += min(grid[i-1][j], grid[i][j-1]);
else if (i>0)
grid[i][j] += grid[i-1][j];
else if (j>0)
grid[i][j] += grid[i][j-1];
}
}
return grid[n-1][m-1];
}
};
322. Coin Change
f[n] = min(f[n-coins[i]], f[n-coins[i+1]], ....) + 1; memo[0] = 0; 如果memo[amount]是0x6fffffff，则表示没走到这过（没发生过更新）返回-1。
class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
int ss = coins.size();
vector<int> memo(amount+1, 0x6fffffff);
memo[0] = 0;
for (int i=1; i<=amount; i++) {
for (int j=0; j<ss; j++) {
if (coins[j]>i) continue;
memo[i] = min(memo[i-coins[j]] + 1, memo[i]);
}
}
return memo[amount] != 0x6fffffff? memo[amount] : -1;
}
};
931. Minimum Falling Path Sum
类似 Minimum Path Sum，
f[x][y] = min(f[x-1][y], f[x-1][y-1], f[x-1][y+1]) + f[x][y]， 可直接在矩阵上操作， 然后判断边缘条件决定转移方程（e.g. y==0时候,忽略f[x-1][y-1]）。
class Solution {
public:
int minFallingPathSum(vector<vector<int>>& A) {
int n = A.size(), m = A[0].size();
for (int x=1; x<n; x++) {
for (int y=0; y<m; y++) {
if (y && m-1-y)
A[x][y] += min(A[x-1][y-1], min(A[x-1][y], A[x-1][y+1]));
else if (y)
A[x][y] += min(A[x-1][y], A[x-1][y-1]);
else if (m-1-y)
A[x][y] += min(A[x-1][y], A[x-1][y+1]);
else
A[x][y] += A[x-1][y];
}
}
return *min_element(A[n-1].begin(), A[n-1].end());
}
};
!983. Minimum Cost For Tickets Medium
f[day]相当于当日若出行，最低总票价
f[day] = min(f[day-1]+ticket_1, f[day-7]+ticket_2, f[day-30]+ticket_2); if f[day] not in days -> f[day] = f[day-1];
class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
int max_day = days[days.size()-1];
vector <int> memo(max_day+1);
int idx = 0; // memo[0] = *min_element(costs.begin(), costs.end());
for (int i=1; i<=max_day; i++) {
if (i<days[idx]) memo[i] = memo[i-1];
else {
if (i>29)
memo[i] = min(min(memo[i-1]+costs[0],
memo[i-7]+costs[1]),
memo[i-30]+costs[2]);
else if (i>6)
memo[i] = min(min(memo[i-1]+costs[0],
memo[i-7]+costs[1]),
costs[2]);
else
memo[i] = min(min(memo[i-1]+costs[0],
costs[1]),
costs[2]);
idx ++;
}
}
return memo[max_day];
}
};
!650. 2 Keys Keyboard Medium
更新都是来自公约数，所以对2n和1n/2循环
dp[n] = min(dp[n], dp[n/cnt] + n/cnt + 1 - 1); AA 复制到 AAAA只要一次粘贴操作
class Solution {
public:
int minSteps(int n) {
if (n == 1) return 0;
vector<int> dp(n+1, 0x6fffffff);
dp[1] = 0, dp[2] = 2;
for (int i=3; i<=n; i++) {
for (int j=i/2; j>0; j--) {
if (i%j) continue;
// AA -> AA AA just need to copy once!
dp[i] = min(dp[i], dp[j] + i/j + 1 - 1);
}
}
return dp[n];
}
};
再简化版 -> f[2]也符合规律
class Solution {
public:
int minSteps(int n) {
vector<int> dp(n+1, 0x6fffffff);
dp[1] = 0;
for (int i=2; i<=n; i++) {
for (int j=i/2; j>0; j--) {
if (i%j) continue;
// AA -> AA AA just need to paste once! and copy also needs one operation. so 1-1;
dp[i] = min(dp[i], dp[j] + i/j + 1 - 1);
}
}
return dp[n];
}
};
279. Perfect Squares Medium
f[n] = min(f[n-sqr_num[i]] + 1, f[n]);
int numSquares(int n) {
if (!n) return 0;
vector<int> dp(n+1, 0x6fffffff), sqr_nums;
for (int i=1; i<=static_cast<int>(sqrt(n)); i++) {
sqr_nums.push_back(pow(i, 2));
}
int ss = sqr_nums.size();
dp[0]=0;
for (int i = 1; i<=n; i++) {
for (int j = 0; j < ss; j++) {
if (sqr_nums[j]>i) break;
dp[i] = min(dp[i-sqr_nums[j]] + 1, dp[i]);
}
}
return dp[n];
}
};
简单优化
找小于等于n的次方for (int i=1; ii<=n; i++) sqr_nums.push_back(ii);
class Solution {
public:
int numSquares(int n) {
// f[n] = min(f[n-sqr_num[i]] + 1, f[n]);
vector<int> dp(n+1, 0x6fffffff), sqr_nums;
for (int i=1; i*i<=n; i++) sqr_nums.push_back(i*i);
int ss = sqr_nums.size();
dp[0]=0;
for (int i = 1; i<=n; i++) {
for (int j = 0; j < ss; j++) {
if (sqr_nums[j]>i) break;
dp[i] = min(dp[i-sqr_nums[j]] + 1, dp[i]);
}
}
return dp[n];
}
};
120. Triangle Medium
类似931. Minimum Falling Path Sum
dp[i][j] = min(dp[i-1][j], dp[i-1][j-1], dp[i-1][j+1]) + dp[i][j], 还要处理边缘的case
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
for (int i=1; i<triangle.size(); i++) {
int ss = triangle[i].size();
for (int j=0; j<ss; j++) {
if (!j)
triangle[i][j] += triangle[i-1][j];
else if (j==ss-1)
triangle[i][j] += triangle[i-1][j-1];
else
triangle[i][j] += min(triangle[i-1][j],
triangle[i-1][j-1]);
}
}
return *min_element(triangle[triangle.size()-1].begin(),
triangle[triangle.size()-1].end());
}
};
1049. Last Stone Weight II Medium
474. Ones and Zeroes Medium
221. Maximal Square Medium
322. Coin Change Medium
1240. Tiling a Rectangle with the Fewest Squares Hard
174. Dungeon Game Hard
871. Minimum Number of Refueling Stops Hard
2.达到目标的方法总数dp(Distinct Ways)
Statement：Given a target find a number of distinct ways to reach the target.
Approach：Sum all possible ways to reach the current state.
routes[i] = routes[i-1] + routes[i-2], ... , + routes[i-k]
70. Climbing Stairs
class Solution {
public:
int climbStairs(int n) {
vector<int> dp(n+1, 1);
for (int i=2; i<=n; i++) {
dp[i] = dp[i-1] + dp[i-2];
}
return dp[n];
}
};
62. Unique Paths
f[i][j] = f[i][j-1]+f[i-1][j];
class Solution {
public:
int uniquePaths(int m, int n) {
// f[i][j] = f[i][j-1]+f[i-1][j];
vector<vector<int>> dp(m, vector<int>(n, 0));
dp[0][0] = 1;
for(int i=0; i<m; i++) {
for (int j=0; j<n; j++) {
if (i && j)
dp[i][j] = dp[i][j-1] + dp[i-1][j];
else if (i)
dp[i][j] = dp[i-1][j];
else if (j)
dp[i][j] = dp[i][j-1];
}
}
return dp[m-1][n-1];
}
};
1155. Number of Dice Rolls With Target Sum
688. Knight Probability in Chessboard
494. Target Sum Medium
377. Combination Sum IV Medium
935. Knight Dialer Medium
1223. Dice Roll Simulation Medium
416. Partition Equal Subset Sum Medium
808. Soup Servings Medium
790. Domino and Tromino Tiling Medium
801. Minimum Swaps To Make Sequences Increasing
673. Number of Longest Increasing Subsequence Medium
63. Unique Paths II Medium
576. Out of Boundary Paths Medium
1269. Number of Ways to Stay in the Same Place After Some Steps Hard
1220. Count Vowels Permutation Hard
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