Update 0746.使用最小花费爬楼梯.md

problems/0746.使用最小花费爬楼梯.md

@@ -399,18 +399,29 @@ impl Solution {
 ### 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];
+#include <math.h>
+int minCostClimbingStairs(int *cost, int costSize) {
+  int dp[costSize + 1];
+  dp[0] = dp[1] = 0;
+  for (int i = 2; i <= costSize; i++) {
+    dp[i] = fmin(dp[i - 2] + cost[i - 2], dp[i - 1] + cost[i - 1]);
+  }
+  return dp[costSize];
+}
+```
 
-    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];
+不使用 dp 数组
+
+```c
+#include <math.h>
+int minCostClimbingStairs(int *cost, int costSize) {
+  int dpBefore = 0, dpAfter = 0;
+  for (int i = 2; i <= costSize; i++) {
+    int dpi = fmin(dpBefore + cost[i - 2], dpAfter + cost[i - 1]);
+    dpBefore = dpAfter;
+    dpAfter = dpi;
+  }
+  return dpAfter;
 }
 ```