LeetCode-Easy-Java-Non-decreasing Array
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改变一个元素的递增数组
题目描述:
Given an array with n
integers, your task is to check if it could become non-decreasing by modifying at most 1
element.
We define an array is non-decreasing if array[i] <= array[i + 1]
holds for every i
(1 <= i < n).
Example 1:
Input: [4,2,3]Output: TrueExplanation: You could modify the first 4
to 1
to get a non-decreasing array.
Example 2:
Input: [4,2,1]Output: FalseExplanation: You can't get a non-decreasing array by modify at most one element.
最初的解决思路就是判断是否出现了两对非递增数,如果出现则返回false,出现了一对非递增数后,则需要判断改变该数后,该数与其前后还是否满足递增关系。
例如 [3,3,2,4],出现的第一对非递增数是[3,3,2,4],可以选择将3变2,成为[3,2,2,4],也可以选择将2变为3[3,3,3,4]。第一种变化后,序列不满足递增,而第二种变化满足。因此变化后的判断逻辑为:
设非递增数对为[a,b],即a>b,若将a变换为b,则判断b是否小于a前一个数;若将b变换为a,则判断a是否大于b后一个数。只有这样才能保证变换后数组仍为递增数组。
代码如下:
public boolean checkPossibility(int[] nums) { int count=0; int position=0; if(nums.length==1){ return true; } for(int i=0;i<nums.length-1;i++){ if(nums[i]<=nums[i+1]){ continue; }else { count++; position=i; if(count>=2){ System.out.println("False"); System.out.println("You can't get a non-decreasing array by modify at most one element"); return false; } } } if(position-1<0||position+2>nums.length-1)//若被变换的数前面也没有,后面也没有则直接返回true{ return true; }else if(position-1>=0&&nums[position-1]<=nums[position+1]){ return true; }else if(position+2<=nums.length-1&&nums[position]<=nums[position+2]){ return true; } return false; }
下面是LeetCode上的优秀代码:
public boolean checkPossibility(int[] nums) { int channum = 0; for(int i=1;i<nums.length;i++){ if(nums[i]<nums[i-1]){ channum++; if(i>1&&nums[i]<nums[i-2]){ //这条语句用于判断b是否小于a前面的数,若小于则将b变换为a nums[i] = nums[i-1]; } } if(channum==2) return false; } return true; }
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