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LeetCode题解,151道题完整版。广告:推荐刷题网站 https://www.lintcode.com/?utm_source=soulmachine

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++inc may be better, although inc++ could AC.

Line 16: return total>=0 ? j+1 : -1 Should be (j+1)%gas.size(), so that we will return 0 instead of N in some cases.

强烈要求使用python实现

提供的代码似乎是题目Plus One的答案,和题目Sqrt(x)没有关系

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77题代码添加cur和k比较条件,可以减少遍历次数 if(cur == k){ result.add(new ArrayList(path)); } // 时间复杂度受k影响,k从2到n,时间复杂度从O(n^2) 到 O(n!) /* n=5 k=2 比较 两次迭代路径,没加cur>k 判断时迭代次数更多。 当n越大越明显。这个便利路径是注释//path.remove(path.size()-1);这个代码后得到的 //123453454552345455345545 //123455 455 345545523455455345545 */ if(cur > k){ return ; }

I have a doubt about the complexity of the recursive method. This should not be n^6. The worst case should be 6^n instead of n^6.