代码随想录额外题目| 数组03 ●34排序数组查首尾位置 ●922按奇偶排序数组II●35搜索插入位置

#34排序数组查首尾位置

medium，我写的:1 暴力

vector<int> searchRange(vector<int>& nums, int target) {int start=-1;int end=-1;for(int i=0;i<nums.size();i++){if(nums[i]==target && start==-1) start=i;if(nums[i]==target && start>-1) end=i;}return {start,end};}

我写的，做了个类似二分搜索的方法：

vector<int> searchRange(vector<int>& nums, int target) {int start=0;int end=nums.size()-1;int res=-1;while(start<=end){//cout<<"start: "<<start<<", end: "<<end<<endl;int mid=(start+end)/2;if(nums[mid]==target){res=mid;break;}if(target<nums[mid]) end=mid-1;if(target>nums[mid]) start=mid+1;}//cout<<res<<endl;start=res; end=res;while(start>=0 && nums[start]==target){start--;}while(end<nums.size() && nums[end]==target){end++;}//if(nums.size()==1) return {0,0};if(res==-1) return{-1,-1};return {start+1,end-1};}

随想录：从两头都做类似二分搜索

     vector<int> searchRange(vector<int>& nums, int target) {int leftBorder = getLeftBorder(nums, target);int rightBorder = getRightBorder(nums, target);// 情况一if (leftBorder == -2 || rightBorder == -2) return {-1, -1};// 情况三if (rightBorder - leftBorder > 1) return {leftBorder + 1, rightBorder - 1};// 情况二return {-1, -1};}int getRightBorder(vector<int>& nums, int target) {int left = 0;int right = nums.size() - 1;int rightBorder = -2; // 记录一下rightBorder没有被赋值的情况while (left <= right) {int middle = left + ((right - left) / 2);if (nums[middle] > target) {right = middle - 1;} else { // 寻找右边界，nums[middle] == target的时候更新leftleft = middle + 1;rightBorder = left;}}return rightBorder;}int getLeftBorder(vector<int>& nums, int target) {int left = 0;int right = nums.size() - 1;int leftBorder = -2; // 记录一下leftBorder没有被赋值的情况while (left <= right) {int middle = left + ((right - left) / 2);if (nums[middle] >= target) { // 寻找左边界right = middle - 1;leftBorder = right;} else {left = middle + 1;}}return leftBorder;}

#922 按奇偶排序数组II

我的解法，有点蠢：

vector<int> sortArrayByParityII(vector<int>& nums) {int cnt=0;vector<int> odd;vector<int> even;for(int i=0;i<nums.size();i++){if(i%2!=0 && nums[i]%2==0) odd.push_back(i);if (i%2==0 && nums[i]%2!=0) even.push_back(i);}for(int i=0;i<odd.size();i++){swap(nums[odd[i]],nums[even[i]]);}return nums;}

inplace解法： 把odd idx放的偶数，给换到even idx放的奇数

注意j是从1开始，而且每轮i，j都是继续增加不回去

空间表现很好，但时间表现很一般

vector<int> sortArrayByParityII(vector<int>& nums) {int j=1;for(int i=0;i<nums.size();i+=2){if(nums[i]%2){while(j<nums.size() && nums[j]%2) j+=2;swap(nums[i],nums[j]);}}return nums;}

#35搜索插入位置

要求复杂度O log n ，我写的：

int searchInsert(vector<int>& nums, int target) {int start=0;int end=nums.size()-1;while(start<=end){int mid=(start+end)/2;//cout<<"mid: "<<mid<<endl;if(nums[mid]==target) return mid;else if(mid==start) {if(start+1<nums.size() && target>nums[start+1]) return start+2;return target>nums[start]?start+1:start;}     if(nums[mid]>target) end=mid-1;if(nums[mid]<target) start=mid+1;}return -1;}

随想录写的：所有情况都能统一变成return end+1或者return start

int searchInsert(vector<int>& nums, int target) {int start=0;int end=nums.size()-1;int mid;while(start<=end){mid=(start+end)/2;if(nums[mid]==target) return mid;if(nums[mid]>target) end=mid-1;if(nums[mid]<target) start=mid+1;}return end+1;}