jump game I && II

Change-Id: I9d598acee6c78a527d1a5f3617c04da06e715335

Author: applewjg

JumpGame.java

@@ -0,0 +1,28 @@
+/*
+ Author:     King, [email protected]
+ Date:       Dec 21, 2014
+ Problem:    Jump Game
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/jump-game/
+ Notes:
+ Given an array of non-negative integers, you are initially positioned at the first index of the array.
+ Each element in the array represents your maximum jump length at that position.
+ Determine if you are able to reach the last index.
+ For example:
+ A = [2,3,1,1,4], return true.
+ A = [3,2,1,0,4], return false.
+
+ Solution: Updated solution: try every reachable index. 
+           Thank to Wenxin Xing for kindly feedback and pointing out my big mistake:)
+ */
+public class Solution {
+    public boolean canJump(int[] A) {
+        int pos = 0, n = A.length;
+        for (int i = 0; i < n; ++i) {
+            if (pos >= i) {
+                pos = Math.max(pos, i + A[i]);
+            }
+        }
+        return pos >= n - 1;
+    }
+}

JumpGameII.java

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+/*
+ Author:     King, [email protected]
+ Date:       Dec 21, 2014
+ Problem:    Jump Game II
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/jump-game-ii/
+ Notes:
+ Given an array of non-negative integers, you are initially positioned at the first index of the array.
+ Each element in the array represents your maximum jump length at that position.
+ Your goal is to reach the last index in the minimum number of jumps.
+ For example:
+ Given array A = [2,3,1,1,4]
+ The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)
+
+ Solution: Jump to the position where we can jump farthest (index + A[index]) next time.
+ */
+public class Solution {
+    public int jump(int[] A) {
+        int n = A.length;
+        int last = 0, cur = 0, res = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i > last) {
+                res++;
+                last = cur;
+            }
+            cur = Math.max(cur, i + A[i]);
+        }
+        return res;
+    }
+}