面试笔试杂项积累-leetcode 161-165

162.162-Find Peak Element-Difficulty: Medium

A peak element is an element that is greater than its neighbors.

Given an input array where num[i] ≠ num[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that num[-1] = num[n] = -∞.

For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.

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Note:

Your solution should be in logarithmic complexity.

思路

找到大于相邻节点的节点

遍历比较

public class Solution {    public int FindPeakElement(int[] nums) {        if (nums.Length < 2)            return 0;      int end = nums.Length - 2;        for (int i = 1; i < end+1; i++)        {            if (nums[i] > nums[i - 1] && nums[i] > nums[i + 1])                return i;            if (nums[end] > nums[end - 1] && nums[end] > nums[end + 1])                return end;            --end;        }        if (nums[0] > nums[1])            return 0;        if (nums[nums.Length - 1] > nums[nums.Length - 2])            return nums.Length - 1;        return 0;    }}

164.164-Maximum Gap-Difficulty: Hard

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Try to solve it in linear time/space.

Return 0 if the array contains less than 2 elements.

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

方法一

思路

求最大相邻数差(排序后)

public class Solution {    public int MaximumGap(int[] nums) {//最大相邻数差                Array.Sort(nums);        int max = 0;        for (int i = 1; i < nums.Length; i++)        {            if (max < nums[i] - nums[i - 1])                max = nums[i] - nums[i - 1];        }        return max;    }}

其他方法

思路

桶排序：http://www.cnblogs.com/ganganloveu/p/4162290.html 

基数排序：https://leetcode.com/discuss/53636/radix-sort-solution-in-java-with-explanation

165.165-Compare Version Numbers-Difficulty: Easy

Compare two version numbers version1 andversion2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the. character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

Here is an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37

思路

比较版本号大小。1.1.3、20.3.4.5这种

每个.与.之间数的比较，相同进入下一个比较，分出大小就有比较结果了

public class Solution {    public int CompareVersion(string version1, string version2) {        if (version1 == null || version2 == null) return 0;        int v1 = 0, v2 = 0;        int i = 0, j = 0;        while (i < version1.Length || j < version2.Length)        {            if (i < version1.Length && version1[i] != '.')            {                v1 = v1 * 10 + (version1[i] - '0');                i++;            }            if (j < version2.Length && version2[j] != '.')            {                v2 = v2 * 10 + (version2[j] - '0');                j++;            }            if (v1 > v2) return 1;            else if (v1 < v2) return -1;        }        return 0;    }}