2019年2月1日

Previous Permutation
摘要： Similar to next permutation, the steps as follow: 1) Find k in the increasing suffix such that nums[k] > nums[k+1], if there is no such element, the p 阅读全文
posted @ 2019-02-01 05:42 一直走在路上 阅读(155) 评论(0) 推荐(0)
Next Permutation
摘要： Leetcode 31 Next permutation [6, 2, 1, 5, 4 3] -> [6, 2, 3, 1, 4, 5] [0, 1, 2] -> [0, 2, 1] [2, 1, 0] -> [0, 1, 2] 1) Find the key k such that nums[k] 阅读全文
posted @ 2019-02-01 04:33 一直走在路上 阅读(160) 评论(0) 推荐(0)

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