数学归纳法
step 1: 验证k0成立
step 2: 验证如果ki成立，那么ki+1也成立
step 3: 联合step1与step2，证明由k0->kn成立

如何解决递推问题

1.确定递推状态

一个函数符号f(x)，外加这个函数符号的含义描述
一般函数所对应的值，就是要求解的值

2.确定递推公式（ki->ki+1）

确定f(x)究竟依赖于哪些f(y)的值

3.分析边界条件（k0）

4.程序实现

递归 || 循环

70. 爬楼梯

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

746. 使用最小花费爬楼梯

class Solution {
public:int minCostClimbingStairs(vector<int>& cost) {int n = cost.size();vector<int> dp(n + 1);cost.push_back(0);dp[0] = cost[0];dp[1] = cost[1];for(int i = 2; i <= n; i++) dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i];return dp[n];}
};

120. 三角形最小路径和

class Solution {
public:int minimumTotal(vector<vector<int>>& triangle) {int n = triangle.size();vector<vector<int>> dp;for(int i = 0; i < 2; i++) dp.push_back(vector<int> (n));for(int i = 0; i < n; i++) dp[(n - 1) % 2][i] = triangle[n - 1][i];for(int i = n - 2; i >= 0; --i){int ind = i % 2;int next_ind = !ind;for(int j = 0; j <= i; j++){dp[ind][j] = min(dp[next_ind][j], dp[next_ind][j + 1]) + triangle[i][j];}}return dp[0][0];}
};

53. 最大子数组和

class Solution {
public:int maxSubArray(vector<int>& nums) {for(int i = 1; i < nums.size(); i++) nums[i] += nums[i - 1];int pre = 0, ans = INT_MIN;for(auto x : nums) {ans = max(x - pre, ans);pre = min(x, pre);}return ans;}
};

122. 买卖股票的最佳时机 II

class Solution {
public:int maxProfit(vector<int>& prices) {int ans = 0;for(int i = 1; i < prices.size(); i++){if(prices[i] > prices[i - 1]) {ans += prices[i] - prices[i - 1];}}return ans;}
};

198. 打家劫舍

class Solution {
public:int rob(vector<int>& nums) {int n = nums.size();vector<vector<int>> dp;for(int i = 0; i < n; i++) dp.push_back(vector<int>(2));dp[0][0] = 0; dp[0][1] = nums[0];for(int i = 1; i < n; i++){dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]);dp[i][1] = dp[i - 1][0] + nums[i];}return max(dp[n - 1][0], dp[n - 1][1]);}
};

322. 零钱兑换

class Solution {
public:int coinChange(vector<int>& coins, int amount) {vector<int> dp(amount + 1);for(int i = 1; i <= amount; i++) dp[i] = -1;for(int i = 1; i <= amount; i++){for(auto x : coins){if(i < x) continue;if(dp[i - x] == -1) continue;if(dp[i] == -1 || dp[i] > dp[i - x] + 1) dp[i] = dp[i - x] + 1;}}return dp[amount];}
};

300. 最长递增子序列

class Solution {
public:int lengthOfLIS(vector<int>& nums) {vector<int> dp(nums.size());int ans = 0;for(int i = 0; i < nums.size(); i++){dp[i] = 1;for(int j = 0; j < i; j++){if(nums[j] >= nums[i]) continue;dp[i] = max(dp[i], dp[j] + 1); }ans = max(dp[i], ans);}return ans;}
};