509.斐波那契数

代码：

class Solution {public int fib(int n) {if (n==0||n==1) {return n;}//1.确认dp数组和下标含义，一维dp数组，第i个数据为第i个数的的fn的值int[] dp = new int[n+1];//2.确认递推公式 fn = f(n-1)+f(n-2)//3.初始化dp数组,dp[0] = 0, dp[1] = 1dp[0] = 0;dp[1] = 1;for (int i = 2; i <= n; i++) {dp[i] = dp[i-1] + dp[i-2];}//4.确认遍历顺序//5.举例推导函数return dp[n];}
}

70.爬楼梯

class Solution {public int climbStairs(int n) {if (n==1) {return 1;}if (n==2) {return 2;}//确认dp数组的含义int[] dp = new int[n+1];//确认递推公式 dp[n] = dp[n-1] + dp[n-2];//确认dp数组的初始化dp[1] = 1;dp[2] = 2;//确认遍历顺序，开始遍历；for (int i = 3; i <= n; i++) {dp[i] = dp[i-1] + dp[i-2];}//举例推导函数return dp[n];}
}

746.使用最小花费爬楼梯

class Solution {public int minCostClimbingStairs(int[] cost) {//确认dp数组；dp[i] 表示跳到第i层楼梯并且再继续往上爬的最小支付费用int[] dp = new int[cost.length];//确认递推公式 dp[i] = Math.min(dp[i-1], dp[i-2]) + cost[i];//dp数组初始化dp[0] = cost[0];dp[1] = cost[1];if (cost.length == 2) {return Math.min(dp[0], dp[1]);}//确认遍历的顺序for (int i = 2; i < cost.length; i++) {dp[i] = Math.min(dp[i-1], dp[i-2]) + cost[i];}//举例推导函数return Math.min(dp[cost.length-1], dp[cost.length-2]);}
}

62.不同路径

class Solution {public int uniquePaths(int m, int n) {/*** 这一题思考了一下，如果想到达终点，那么就只能从终点的上一个位置或者从终点的左侧位置到达，* 那么可以推导出 dp[m][n] = dp[m-1][n] + dp[m][n-1];**/if (m == 1||n==1) {return 1;}//确定dp数组int[][] dp = new int[m+1][n+1];//确认递推公式，dp[m][n] = dp[m-1][n] + dp[m][n-1];//dp数组的初始化,先给第二列和第二行的数据都调整为1for (int i = 1; i < m+1; i++) {dp[i][1] = 1;}for (int i = 1; i < n+1; i++) {dp[1][i] = 1;}//确认dp数组的遍历顺序,从第二列第二行开始遍历整个表格，for (int i = 2; i < m+1; i++) {for (int j = 2; j < n+1; j++) {dp[i][j] = dp[i-1][j] + dp[i][j-1];}}//举例推导函数return dp[m][n];}
}

63.不同路径II

class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;if (obstacleGrid[0][0] == 1) {return 0;}if (obstacleGrid[m-1][n-1] == 1) {return 0;}if (m==1||n==1) {for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (obstacleGrid[i][j] == 1) {return 0;}}}return 1;}//确认dp数组int[][] dp = new int[m+1][n+1];//确认递推公式 递推函数依然为 dp[i][j] = dp[i][j-1] + dp[i-1][j];//初始化递推数组dp[1][1] = 1;//确认遍历顺序for (int i = 1; i <= m; i++) {for (int j = 1; j <= n; j++) {if (i==1&&j==1) continue;if (obstacleGrid[i-1][j-1] == 1) {dp[i][j] = 0;} else {dp[i][j] = dp[i-1][j] + dp[i][j-1];}}}//举例推导函数return dp[m][n];}
}

343.整数拆分

class Solution {public int integerBreak(int n) {if (n == 2) {return 1;}//确认dp数组含义int[] dp = new int[n+1];//确认递推公式, dp[i] = Math.max(dp[i], Math.max(dp[j]*(i-j), (i-j) * j));//dp数组初始化dp[2] = 1;//确认dp数组的遍历顺序for (int i = 3; i <= n; i++) {for (int j = 1; j < i-1; j++) {dp[i] = Math.max(dp[i], Math.max(dp[j] * (i-j), (i-j) * j));}}//举例推导结果函数return dp[n];}
}