java - 【算法题】给定int数组，移除不超过一个元素后，判断是否存在自增序列

问题描述

没什么思路啊，题目如下Given a sequence of integers as an array, determine whether it is possible to obtain a strictly increasing sequence by removing no more than one element from the array.

Example

For sequence = [1, 3, 2, 1], the output should bealmostIncreasingSequence(sequence) = false;

There is no one element in this array that can be removed in order to get a strictly increasing sequence.

For sequence = [1, 3, 2], the output should bealmostIncreasingSequence(sequence) = true.

You can remove 3 from the array to get the strictly increasing sequence [1, 2]. Alternately, you can remove 2 to get the strictly increasing sequence [1, 3].

Input/Output

[time limit] 4000ms (js)[input] array.integer sequence

Guaranteed constraints:2 ≤ sequence.length ≤ 105,-105 ≤ sequence[i] ≤ 105.

[output] boolean

Return true if it is possible to remove one element from the array in order to get a strictly increasing sequence, otherwise return false.

有个思路：2层循环，第一循环移除元素，第二层循环判断移除这个元素后是否有自增序列。

问题解答

提供一个思路

作出逐差数组: 如 a=[1,3,2,1]，逐差后得 [2,-1,-1]

所谓删除一个元素，即在在逐差数组中去头或去尾，或把相邻两个相加合并成一个元素。

因此，若逐差数组中有多于一个负数，则不行; 若无负数，则可以; 否则对惟一的负数作以上操作，若其能被删掉或被合并成正数，则可以

这样一来，时间复杂度可以降到 O(n)

可以在 O(n) 时间做到:

对每个相邻的 [a, b]，判断是否 a >= b。这样的数对破坏严格递增性。如果这样的数对超过一个，返回false。如果一个也没有，返回true。

如果1中只有一对 [a0, b0]，判断 '移除a0或b0后是否还是递增' 并返回

结果是对的，但是超过规定的时间了，有更好的方法吗？

function almostIncreasingSequence(sequence) { var iscan = false; var is = true; var temp for(var i=0;i<sequence.length;i++){is = true;temp = sequence.slice(0,i).concat(sequence.slice(i+1));for(var j=0;j+1<temp.length;j++){ if(temp[j] <= temp[j+1]){is = false;break; }}if(is){ iscan=true; break;} } return iscan;}

时间复杂度为O(n)的方法

boolean almostIncreasingSequence(int[] sequence) { if(sequence.length<=2){return true; } //找出逆序的数的index int count = 0; int biggerIndex = 0; int smallerIndex = 0; boolean isHave = true; for(int i=0;i+1<sequence.length;i++){//如果找到2组逆序，直接返回falseif(count>1){ isHave = false;}if(sequence[i]>=sequence[i+1]){ count ++; biggerIndex = i; smallerIndex = i+1;} }//分别判断当移除2个数后剩下的数组是不是自增的 for(int i=0;i+2<sequence.length;i++){int cur = i;int next = i+1;if(i==biggerIndex){ continue;}if(i+1==biggerIndex){ next = i+2;}if(sequence[cur]>=sequence[next]){ isHave = false;} } if(isHave){return isHave; }else{isHave = true; } for(int i=0;i+2<sequence.length;i++){int cur = i;int next = i+1;if(i==smallerIndex){ continue;}if(i+1==smallerIndex){ next = i+2;}if(sequence[cur]>=sequence[next]){ isHave = false; } } return isHave;}回答4：

这个是不是统计逆序数的个数？