leetcode笔记—关于动态规划

动态规划的思路：找到DP（1），DP（2）,DP（n）与DP（n-1）、DP(n-2)的关系

</pre><pre name="code" class="cpp">

1

class Solution {public:    int integerBreak(int n) {        if(n==2)    return 1;          if(n==3)    return 2;          vector<int> Dp(n+1,0);          Dp[0] = 1;Dp[1] = 1;          Dp[2] = 2;Dp[3] = 3;          for(int i=4; i<=n; ++i){                  Dp[i] = max(Dp[i/2]*Dp[i-i/2],max(Dp[i-3]*Dp[3],Dp[i-2]*Dp[2]));//两边尽量相等时乘积最大          }          return Dp[n];      }};

2.Coin Change

class Solution {public:const int INF = 0x7ffffffe;    int coinChange(vector<int>& coins, int amount) {        // 无效输入的处理        if (amount == 0)            return 0;        if (coins.size() == 0)            return -1;                    vector<int> dp(amount+1,0);        for (int i = 1; i <= amount; i++) {            int min1= INF;            for (int j = 0; j < coins.size(); j++) {                if (i >= coins[j] && dp[i - coins[j]] != -1)                    min1= min(min1, dp[i - coins[j]] + 1);            }                        // 根据min的值判断是否能兑换            dp[i] = min1 == INF ? -1 : min1;        }        return dp[amount];    }};

3. Climbing Stairs

有n个台阶每次可以走一个或者两个，有多少种走法？

DP（1）=1，DP（2）=2，DP（n）=DP（n-1）+DP（n-2）

class Solution {public:    int climbStairs(int n) {        vector<int> dp(n);        if(n<=2) return n;        else        {            dp[0]=1;            dp[1]=2;            for(int i=2;i<n;i++)            {                dp[i]=dp[i-1]+dp[i-2];                            }            return dp[n-1];        }    }};

3.Unique Paths

开一个f[m][n]的数组，f[i][j] = f[i-1][j] + f[i][j-1]，空间时间复杂度O(m*n)。用滚动数组空间复杂度可降为O(n)

class Solution {public:    int uniquePaths(int m, int n) {        vector<vector<int> > f(m, vector<int>(n));        for(int i=0;i<m;i++)        {            f[i][0]=1;        }        for(int i=0;i<n;i++)        {            f[0][i]=1;        }        for(int i=1;i<m;i++)          for(int j=1;j<n;j++)             f[i][j]=f[i-1][j]+f[i][j-1];        return f[m-1][n-1];    }};

4.Unique Paths 2

有障碍

[  [0,0,0],  [0,1,0],  [0,0,0]]

class Solution {public:    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {        vector<vector<int> > f(obstacleGrid.size(), vector<int>(obstacleGrid[0].size()));        f[0][0]=obstacleGrid[0][0]==1 ? 0:1;        int m=obstacleGrid.size();        int n=obstacleGrid[0].size();        for(int i=1;i<m;i++)        {            f[i][0]=obstacleGrid[i][0]==1 ? 0:f[i-1][0];        }        for(int i=1;i<n;i++)        {            f[0][i]=obstacleGrid[0][i]==1 ? 0:f[0][i-1];        }        for(int i=1;i<m;i++)          for(int j=1;j<n;j++)           f[i][j]=obstacleGrid[i][j]==1 ? 0:f[i-1][j]+f[i][j-1];    return f[m-1][n-1];    }};

4.Burst Balloons 重点 难点

class Solution {public:    int maxCoins(vector<int>& nums) {        int n=nums.size();        nums.insert(nums.begin(), 1);        nums.push_back(1);        vector<vector<int>> dp(nums.size(), vector<int>(nums.size(), 0));       for (int len = 1; len <= n; ++len)            for (int left = 1; left <= n - len + 1; ++left) {                int right = left + len - 1;                for (int k = left; k <= right; ++k)                    dp[left][right] = max(dp[left][right], nums[left-1]*nums[k]*nums[right+1] + dp[left][k-1] + dp[k+1][right]);            }        return dp[1][n];    }};