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Given a permutation which contains no repeated number, find its index in all the permutations of these numbers, which are ordered in lexicographical order. The index begins at 1.
Given [1,2,4], return 1.
Thought:
1)Initial thought:
- sort all numbers in ascending order;
- index+=index in sorted list * factor;
- factor /=(A.length-i);
Time complexity: sort- (nlogn) + loop(N * N-remove element from list) ===> o(N^2)
Space complexity: O(N)
public long permutationIndex(int[] A) { // Write your code here if(A==null||A.length==0){ return 0; } List<Integer> li=new ArrayList<Integer>(); for(int i=0;i<A.length;i++){ li.add(A[i]); } Collections.sort(li); long res=1; long factor=1; for(int i=2;i<A.length;i++){ factor*=i; } int i=0; while(li.size()>1){ int cur=li.indexOf(A[i++]); res+=cur*factor; li.remove(cur); factor/=(A.length-i); } return res; }
2) Optimized Solution
Original source: http://algorithm.yuanbin.me/zh-cn/exhaustive_search/previous_permuation.html
- for each element at index i from outer loop,we check elements right after it. If current element > any following element,rank++;
- index+= rank(number of elements smaller than this number)*factor(number of combinations for each number);
- factor*=length-i;
Time complexity:O(N^2);
Space complexity:O(1);
public long permutationIndex(int[] A) { // Write your code here if(A==null||A.length==0){ return 0; } long res=1; long factor=1; for(int i=A.length-1;i>=0;i--){ int count=0; for(int j=i+1;j<A.length;j++){ if(A[i]>A[j]){ count++; } } res+=count*factor; factor*=(A.length-i); } return res; }
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原文地址：http://www.cnblogs.com/fifi043/p/4993657.html
