Merge pull request #1014 from KingArthur0205/remote

添加 0509.斐波那契数.md, 0070.爬楼梯.md, 0746.使用最小花费爬楼梯.md, 0062.不同路径.md, 0063.不同路径II.md  C语言版本

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;
+}
+```
 
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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];
+}
+```
 
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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];
+}
+```
+
 
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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);
+}
+```
 
 
 -----------------------

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];
+}
+```
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 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>