Merge branch 'master' of github.com:youngyangyang04/leetcode-master

problems/0121.买卖股票的最佳时机.md

@@ -308,7 +308,7 @@ class Solution:
 class Solution:
     def maxProfit(self, prices: List[int]) -> int:
         length = len(prices)
-        if len == 0:
+        if length == 0:
             return 0
         dp = [[0] * 2 for _ in range(length)]
         dp[0][0] = -prices[0]

problems/0134.加油站.md

@@ -345,6 +345,37 @@ class Solution:
 ```
 
 ### Go 
+
+贪心算法（方法一）
+```go
+func canCompleteCircuit(gas []int, cost []int) int {
+    curSum := 0
+    min := math.MaxInt64
+    for i := 0; i < len(gas); i++ {
+        rest := gas[i] - cost[i]
+        curSum += rest
+        if curSum < min {
+            min = curSum
+        }
+    }
+    if curSum < 0 {
+        return -1
+    }
+    if min >= 0 {
+        return 0
+    }
+    for i := len(gas) - 1; i > 0; i-- {
+        rest := gas[i] - cost[i]
+        min += rest
+        if min >= 0 {
+            return i
+        }
+    }
+    return -1
+}
+```
+
+贪心算法（方法二）
 ```go
 func canCompleteCircuit(gas []int, cost []int) int {
 	curSum := 0

problems/0455.分发饼干.md

@@ -226,21 +226,36 @@ class Solution:
 ```
 
 ### Go
-```golang
-//排序后，局部最优
+
+版本一 大饼干优先
+```Go
 func findContentChildren(g []int, s []int) int {
-  sort.Ints(g)
-  sort.Ints(s)
-
-  // 从小到大
-  child := 0
-  for sIdx := 0; child < len(g) && sIdx < len(s); sIdx++ {
-    if s[sIdx] >= g[child] {//如果饼干的大小大于或等于孩子的为空则给与，否则不给予，继续寻找选一个饼干是否符合
-      child++
+    sort.Ints(g)
+    sort.Ints(s)
+    index := len(s) - 1
+    result := 0
+    for i := len(g) - 1; i >= 0; i-- {
+        if index >= 0 && s[index] >= g[i] {
+            result++
+            index--
+        }
     }
-  }
+    return result
+}
+```
 
-  return child
+版本二 小饼干优先
+```Go
+func findContentChildren(g []int, s []int) int {
+    sort.Ints(g)
+    sort.Ints(s)
+    index := 0
+    for i := 0; i < len(s); i++ {
+        if index < len(g) && g[index] <= s[i] {
+            index++
+        }
+    }
+    return index
 }
 ```
 

problems/0474.一和零.md

@@ -615,7 +615,7 @@ impl Solution {
     }
 }
 ```
-## C
+### C
 
 ```c
 #define max(a, b) ((a) > (b) ? (a) : (b))

problems/0494.目标和.md

@@ -585,7 +585,7 @@ impl Solution {
     }
 }
 ```
-## C
+### C
 
 ```c
 int getSum(int * nums, int numsSize){

problems/kamacoder/0103.水流问题.md

@@ -355,6 +355,62 @@ public class Main {
 ```
 
 ### Python
+```Python
+first = set()
+second = set()
+directions = [[-1, 0], [0, 1], [1, 0], [0, -1]]
+
+def dfs(i, j, graph, visited, side):
+    if visited[i][j]:
+        return
+    
+    visited[i][j] = True
+    side.add((i, j))
+    
+    for x, y in directions:
+        new_x = i + x
+        new_y = j + y
+        if (
+            0 <= new_x < len(graph)
+            and 0 <= new_y < len(graph[0])
+            and int(graph[new_x][new_y]) >= int(graph[i][j])
+        ):
+            dfs(new_x, new_y, graph, visited, side)
+
+def main():
+    global first
+    global second
+    
+    N, M = map(int, input().strip().split())
+    graph = []
+    for _ in range(N):
+        row = input().strip().split()
+        graph.append(row)
+    
+    # 是否可到达第一边界
+    visited = [[False] * M for _ in range(N)]
+    for i in range(M):
+        dfs(0, i, graph, visited, first)
+    for i in range(N):
+        dfs(i, 0, graph, visited, first)
+    
+    # 是否可到达第二边界
+    visited = [[False] * M for _ in range(N)]
+    for i in range(M):
+        dfs(N - 1, i, graph, visited, second)
+    for i in range(N):
+        dfs(i, M - 1, graph, visited, second)
+
+    # 可到达第一边界和第二边界
+    res = first & second
+    
+    for x, y in res:
+        print(f"{x} {y}")
+    
+    
+if __name__ == "__main__":
+    main()
+```
 
 ### Go
 

problems/周总结/20201017二叉树周末总结.md

@@ -40,7 +40,7 @@
 
 * 陷阱二
 
-在一个有序序列求最值的时候，不要定义一个全局遍历，然后遍历序列更新全局变量求最值。因为最值可能就是int 或者 longlong的最小值。
+在一个有序序列求最值的时候，不要定义一个全局变量，然后遍历序列更新全局变量求最值。因为最值可能就是int 或者 longlong的最小值。
 
 推荐要通过前一个数值（pre）和后一个数值比较（cur），得出最值。