本质上感觉是一个相加的问题，状态的转换，由前一种状态推至下一种状态
509. 斐波那契数
较为简单
746. 使用最小花费爬楼梯
62. 不同路径
一开始写的时候被吓到了，但是发现听完一半之后再写还是比较容易的
对于我而言主要是找到逻辑，
class Solution {public int uniquePaths(int m, int n) {if (m <= 1 || n <=1){return 1;}int[][] result = new int[m][n];result[0][0] = 0;result[1][0] = 1;result[0][1] = 1;for (int i = 0;i < m;i++){for (int j = 0;j < n; j++){if (i > 0 && j > 0){result[i][j] = result[i][j-1] + result[i-1][j];}else if (i == 0 && j > 0 && j!=1){result[i][j] = result[i][j-1];}else if (j == 0 && i > 0 && i!=1){result[i][j] = result[i-1][j];}}}return result[m-1][n-1];
}
}63. 不同路径 II
秒杀，和前者逻辑差不多，要注意初始化就好
class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;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;}int[][] result = new int[m][n];if (obstacleGrid[0][0] == 1){return 0;}else{result[0][0] = 0;}if (obstacleGrid[1][0] == 1){result[1][0] = 0;}else{result[1][0] = 1;}if (obstacleGrid[0][1] == 1){result[0][1] = 0;}else{result[0][1] = 1;}for (int i = 0;i < m;i++){for (int j = 0;j < n; j++){if (obstacleGrid[i][j] == 1){result[i][j] = 0;}else if (i > 0 && j > 0){result[i][j] = result[i][j-1] + result[i-1][j];}else if (i == 0 && j > 0 && j!=1){result[i][j] = result[i][j-1];}else if (j == 0 && i > 0 && i!=1){result[i][j] = result[i-1][j];}}}return result[m-1][n-1];
}
}别人的写法，比较nb
class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int m = obstacleGrid.length;int n = obstacleGrid[0].length;int[][] dp = new int[m][n];//如果在起点或终点出现了障碍，直接返回0if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {return 0;}for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {dp[i][0] = 1;}for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {dp[0][j] = 1;}for (int i = 1; i < m; i++) {for (int j = 1; j < n; j++) {dp[i][j] = (obstacleGrid[i][j] == 0) ? dp[i - 1][j] + dp[i][j - 1] : 0;}}return dp[m - 1][n - 1];}
}343. 整数拆分
再写吧。。累了
96. 不同的二叉搜索树
416. 分割等和子集
背包问题 - 我真心实意的感觉这个像玄学
明明其实不太明白呢，但神不知鬼不觉的就写出来了，我的天，，，，，
为什么要求max呢？？？？我真的不理解
class Solution {public boolean canPartition(int[] nums) {int sum =0;for (int i =0;i < nums.length;i++){sum += nums[i];}if (sum % 2 != 0){return false;}else{sum = sum/2;}int dp[][] = new int[nums.length][sum + 1];for (int j = 0;j < sum + 1;j++){if (j >= nums[0]){dp[0][j] = nums[0];}}for (int i = 1;i < nums.length;i++){for (int j = 1;j < sum + 1;j++){if (j < nums[i]) dp[i][j] = dp[i - 1][j];else dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - nums[i]] + nums[i]); }}if(dp[nums.length - 1][sum] == sum){return true;}else{return false;}}
}
)
