1.碰到next_permutation(permutation:序列的意思) 今天在TC上碰到一道简单题(SRM531 - Division Two - Level One),是求给定数组不按升序排列的最小字典序列(Sequence of numbers A is lexicographically smaller than B if A contains a smaller number on the first position on which they differ). 解法很简单,就
描述 Given a collection of distinct integers, return all possible permutations. Example: Input: [1,2,3] Output: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ] 思路一:递归 利用STL的函数swap()来交换数组的元素 class Solution { public: vector<vector<int>> p
这两个函数都包含在algorithm库中.STL提供了两个用来计算排列组合关系的算法,分别是next_permutation和prev_permutation. 一.函数原型 首先我们来看看这两个函数的函数原型: next_permutation: template< class BidirIt >bool next_permutation( BidirIt first, BidirIt last ); template< class BidirIt, class Compare >