diff --git a/problems/0062.不同路径.md b/problems/0062.不同路径.md index e209abb1..efa85a03 100644 --- a/problems/0062.不同路径.md +++ b/problems/0062.不同路径.md @@ -347,6 +347,42 @@ var uniquePaths = function(m, n) { }; ``` +### C +```c +//初始化dp数组 +int **initDP(int m, int n) { + //动态开辟dp数组 + int **dp = (int**)malloc(sizeof(int *) * m); + int i, j; + for(i = 0; i < m; ++i) { + dp[i] = (int *)malloc(sizeof(int) * n); + } + + //从0,0到i,0只有一种走法,所以dp[i][0]都是1,同理dp[0][j]也是1 + for(i = 0; i < m; ++i) + dp[i][0] = 1; + for(j = 0; j < n; ++j) + dp[0][j] = 1; + return dp; +} + +int uniquePaths(int m, int n){ + //dp数组,dp[i][j]代表从dp[0][0]到dp[i][j]有几种走法 + int **dp = initDP(m, n); + + int i, j; + //到达dp[i][j]只能从dp[i-1][j]和dp[i][j-1]出发 + //dp[i][j] = dp[i-1][j] + dp[i][j-1] + for(i = 1; i < m; ++i) { + for(j = 1; j < n; ++j) { + dp[i][j] = dp[i-1][j] + dp[i][j-1]; + } + } + int result = dp[m-1][n-1]; + free(dp); + return result; +} +``` -----------------------
diff --git a/problems/0063.不同路径II.md b/problems/0063.不同路径II.md index 490b6b5c..6f405d6a 100644 --- a/problems/0063.不同路径II.md +++ b/problems/0063.不同路径II.md @@ -333,6 +333,60 @@ var uniquePathsWithObstacles = function(obstacleGrid) { }; ``` +C +```c +//初始化dp数组 +int **initDP(int m, int n, int** obstacleGrid) { + int **dp = (int**)malloc(sizeof(int*) * m); + int i, j; + //初始化每一行数组 + for(i = 0; i < m; ++i) { + dp[i] = (int*)malloc(sizeof(int) * n); + } + + //先将第一行第一列设为0 + for(i = 0; i < m; ++i) { + dp[i][0] = 0; + } + for(j = 0; j < n; ++j) { + dp[0][j] = 0; + } + + //若碰到障碍,之后的都走不了。退出循环 + for(i = 0; i < m; ++i) { + if(obstacleGrid[i][0]) { + break; + } + dp[i][0] = 1; + } + for(j = 0; j < n; ++j) { + if(obstacleGrid[0][j]) + break; + dp[0][j] = 1; + } + return dp; +} + +int uniquePathsWithObstacles(int** obstacleGrid, int obstacleGridSize, int* obstacleGridColSize){ + int m = obstacleGridSize, n = *obstacleGridColSize; + //初始化dp数组 + int **dp = initDP(m, n, obstacleGrid); + + int i, j; + for(i = 1; i < m; ++i) { + for(j = 1; j < n; ++j) { + //若当前i,j位置有障碍 + if(obstacleGrid[i][j]) + //路线不同 + dp[i][j] = 0; + else + dp[i][j] = dp[i-1][j] + dp[i][j-1]; + } + } + //返回最后终点的路径个数 + return dp[m-1][n-1]; +} +``` -----------------------
diff --git a/problems/0070.爬楼梯.md b/problems/0070.爬楼梯.md index 97db670a..7f47a991 100644 --- a/problems/0070.爬楼梯.md +++ b/problems/0070.爬楼梯.md @@ -296,6 +296,48 @@ var climbStairs = function(n) { }; ``` +### C +```c +int climbStairs(int n){ + //若n<=2,返回n + if(n <= 2) + return n; + //初始化dp数组,数组大小为n+1 + int *dp = (int *)malloc(sizeof(int) * (n + 1)); + dp[0] = 0, dp[1] = 1, dp[2] = 2; + + //从前向后遍历数组,dp[i] = dp[i-1] + dp[i-2] + int i; + for(i = 3; i <= n; ++i) { + dp[i] = dp[i - 1] + dp[i - 2]; + } + //返回dp[n] + return dp[n]; +} +``` + +优化空间复杂度: +```c +int climbStairs(int n){ + //若n<=2,返回n + if(n <= 2) + return n; + //初始化dp数组,数组大小为3 + int *dp = (int *)malloc(sizeof(int) * 3); + dp[1] = 1, dp[2] = 2; + + //只记录前面两个台阶的状态 + int i; + for(i = 3; i <= n; ++i) { + int sum = dp[1] + dp[2]; + dp[1] = dp[2]; + dp[2] = sum; + } + //返回dp[2] + return dp[2]; +} +``` + -----------------------
diff --git a/problems/0509.斐波那契数.md b/problems/0509.斐波那契数.md index d8020d83..c6ce76c0 100644 --- a/problems/0509.斐波那契数.md +++ b/problems/0509.斐波那契数.md @@ -245,6 +245,38 @@ var fib = function(n) { }; ``` +### C +动态规划: +```c +int fib(int n){ + //当n <= 1时,返回n + if(n <= 1) + return n; + //动态开辟一个int数组,大小为n+1 + int *dp = (int *)malloc(sizeof(int) * (n + 1)); + //设置0号位为0,1号为为1 + dp[0] = 0; + dp[1] = 1; + + //从前向后遍历数组(i=2; i <= n; ++i),下标为n时的元素为dp[i-1] + dp[i-2] + int i; + for(i = 2; i <= n; ++i) { + dp[i] = dp[i - 1] + dp[i - 2]; + } + return dp[n]; +} +``` + +递归实现: +```c +int fib(int n){ + //若n小于等于1,返回n + if(n <= 1) + return n; + //否则返回fib(n-1) + fib(n-2) + return fib(n-1) + fib(n-2); +} +``` ----------------------- diff --git a/problems/0746.使用最小花费爬楼梯.md b/problems/0746.使用最小花费爬楼梯.md index 0009f06c..e94e4d24 100644 --- a/problems/0746.使用最小花费爬楼梯.md +++ b/problems/0746.使用最小花费爬楼梯.md @@ -266,5 +266,21 @@ var minCostClimbingStairs = function(cost) { }; ``` +### C +```c +int minCostClimbingStairs(int* cost, int costSize){ + //开辟dp数组,大小为costSize + int *dp = (int *)malloc(sizeof(int) * costSize); + //初始化dp[0] = cost[0], dp[1] = cost[1] + dp[0] = cost[0], dp[1] = cost[1]; + + int i; + for(i = 2; i < costSize; ++i) { + dp[i] = (dp[i-1] < dp[i-2] ? dp[i-1] : dp[i-2]) + cost[i]; + } + //选出倒数2层楼梯中较小的 + return dp[i-1] < dp[i-2] ? dp[i-1] : dp[i-2]; +} +``` -----------------------