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This problem has a very concise solution in this link, just 4-lines. The code is rewritten below.
1 class Solution { 2 public: 3 bool canJump(vector<int>& nums) { 4 int i = 0, n = nums.size(); 5 for (int r = 0; i < n && i <= r; i++) 6 r = max(r, i + nums[i]); 7 return i == n; 8 } 9 };
The above code can be further optimized by avoiding unnecessary checks. Once we are able to reach the last position, we are already done.
1 class Solution { 2 public: 3 bool canJump(vector<int>& nums) { 4 int i = 0, r = 0, n = nums.size(); 5 for (; i < n && i <= r && r < n - 1; i++) 6 r = max(r, i + nums[i]); 7 return r >= n - 1; 8 } 9 };
Or we can rewrite the code a bit longer to make the idea obvious.
1 class Solution { 2 public: 3 bool canJump(int A[], int n) { 4 int start = 0, end = 0; 5 while(start <= end && end < n - 1) { 6 end = max(end, start + A[start]); 7 start++; 8 } 9 return end >= n - 1; 10 } 11 };
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4649395.html
