[pull] master from youngyangyang04:master

problems/0063.不同路径II.md

@@ -210,24 +210,54 @@ public:
 ```java
 class Solution {
     public int uniquePathsWithObstacles(int[][] obstacleGrid) {
-        int n = obstacleGrid.length, m = obstacleGrid[0].length;
-        int[][] dp = new int[n][m];
-
-        for (int i = 0; i < m; i++) {
-	    if (obstacleGrid[0][i] == 1) break; //一旦遇到障碍，后续都到不了
-	    dp[0][i] = 1;
+        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0
+        if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) {
+            return 0;
         }
-        for (int i = 0; i < n; i++) {
-	    if (obstacleGrid[i][0] == 1) break; ////一旦遇到障碍，后续都到不了
-	    dp[i][0] = 1;
+
+        for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) {
+            dp[i][0] = 1;
         }
-        for (int i = 1; i < n; i++) {
-            for (int j = 1; j < m; j++) {
-                if (obstacleGrid[i][j] == 1) continue;
-                dp[i][j] = dp[i - 1][j] + dp[i][j - 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];
+    }
+}
+```
+
+```java
+// 空间优化版本
+class Solution {
+    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
+        int m = obstacleGrid.length;
+        int n = obstacleGrid[0].length;
+        int[] dp = new int[n];
+
+        for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) {
+            dp[j] = 1;
+        }
+
+        for (int i = 1; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if (obstacleGrid[i][j] == 1) {
+                    dp[j] = 0;
+                } else if (j != 0) {
+                    dp[j] += dp[j - 1];
+                }
             }
         }
-        return dp[n - 1][m - 1];
+        return dp[n - 1];
     }
 }
 ```