1. Problem Descriptions:Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value.If target is not found in the array, return [-1…
Find First and Last Position of Element in Sorted Array - LeetCode 注意点 nums可能为空 时间复杂度为O(logn) 解法 解法一:最普通的二分搜索,先找到一个target,然后向两边拓展。 class Solution { public: int binarySearch(vector<int>& nums, int target) { int left = 0,right = nums.size()-1;...
target,false) -11516returntargetRange17}1819//返回 `target`应该是最左边(或最右边)的索引20//通过二进制搜索插入排序数组'nums'.21func extremeInsertionIndex(_ nums: [Int], _ target: Int,_ left:Bool) ->Int22{23varlo:Int =024varhi:Int =nums.count2526while(lo <hi)27{28varmid...
1) Code: // 方案3 “二分查找” // 参考: // 1)https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/zai-pai-xu-shu-zu-zhong-cha-zhao-yuan-su-de-di-3-4/ const binarySearch = (nums, target, lower) => { let left = 0, right = num...
[LeetCode] 34. Find First and Last Position of Element in Sorted Array 2019-11-04 12:18 − Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value. If target is not found i... CNoodle 0 521 Head First设计...
[LeetCode] 34. Find First and Last Position of Element in Sorted Array 2019-11-04 12:18 −Given an array of integers nums sorted in non-decreasing order, find the starting and ending position of a given target value. If target is not found i... ...
()-1;intindex=-1;while(left<=right){mid=(left+right)/2;inttmp=nums[mid];if(tmp<target){left=mid+1;}elseif(tmp>target){right=mid-1;}else{index=mid;break;}}if(index==-1){inta[2]={-1,-1};vector<int>b(a,a+2);returnb;}vector<int>res;intl=rangeSearch(nums,0,index,...
1 二分法查找的时候,还是要用while l<=r,这样写可以cover输入只有一个元素的情况 2 这里list和string都有count函数,但就是前后要对应上 3 很多...
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Leetcode知识点: list isalpha join 推导式 pop 交换元素 class Solution: def reverseOnlyLetters(self, s: str) -> str: # 一、基础语法 x = list(s) # 转换成列表 i, j = 0, len(s) - 1 while i < j: a, b = x[i].isalpha(),x[j].isalpha() if not a: i += 1 if not b: ...