代码随想录算法训练营打卡day1 |704. 二分查找，27. 移除元素

一、LeetCode 704 二分查找

题目链接：704.二分查找

解法一：左闭右闭

class Solution {public int search(int[] nums, int target) {int left = 0, right = nums.length-1;while(left <= right){ //边界处理，左闭右闭int mid = left + (right-left)/2;if(nums[mid] < target){        //target在右，左指针后移left = mid+1;}else if(nums[mid] > target){ //target在左，右指针前移right = mid-1;}else{return mid;}}return -1;}
}

解法二：左闭右开

class Solution {public int search(int[] nums, int target) {int left = 0, right = nums.length;while(left < right){ //边界处理，左闭右开int mid = left + (right-left)/2;if(nums[mid] < target){        //target在右，左指针后移left = mid+1;}else if(nums[mid] > target){ //target在左，右指针前移right = mid;}else{return mid;}}return -1;}
}

扩展题目

35.搜索插入位置

思路：利用二分法找到插入位置：若存在，返回下标；若不存在，退出while循环，返回下标left（left = right+1）。

代码如下：

class Solution {public int searchInsert(int[] nums, int target) {int left = 0, right = nums.length-1;while(left <= right){int mid = left + (right-left)/2;if(nums[mid] < target){left = mid+1;}else if(nums[mid] > target){right = mid-1;}else{return mid;}}return left;// left = right+1}
}

34.在排序数组中查找第一个和最后一个元素的位置

思路：排序数组-可以使用二分。利用二分法找到任意该元素的位置；向左和向右找到边界位置。

代码：

class Solution {public int[] searchRange(int[] nums, int target) {int left = 0, right = nums.length-1;int[] ans = new int[2];int mid = 0;//先使用二分法找到target值while(left <= right){mid = left + (right-left)/2;if(nums[mid] < target){left = mid+1;}else if(nums[mid] > target){right = mid-1;}else{break;}}//确定起始点和终点if(left > right){ans[0] = -1;ans[1] = -1;}else{int index = mid;while(index >= 0 && nums[index] == target){index--;}ans[0] = index+1; //返回下标index = mid;while(index < nums.length && nums[index] == target){index++;}ans[1] = index-1;}return ans;}
}

二、LeetCode 27 移除元素

题目链接：27.移除元素

思路：设置双指针，一次遍历过程中将指定元素替换到新的数组长度之外。

class Solution {public int removeElement(int[] nums, int val) {int n = nums.length;int left = 0, right = n-1;//双指针遍历while(left <= right){if(nums[left] == val){int temp = nums[left];nums[left] = nums[right];nums[right] = temp;left--; //回溯right--;}left++;}return right+1;}
}