第一天 / 704. 二分查找 / 27. 移除元素 / 977. 有序数组的平方

704. 二分查找

• 左闭右闭

      class Solution {public:int search(vector<int>& nums, int target) {int left = 0;int right = nums.size() - 1; // 定义target在左闭右闭的区间里，[left, right]while (left <= right) { // 当left==right，区间[left, right]依然有效，所以用 <=int middle = left + ((right - left) / 2);// 防止溢出 等同于(left + right)/2if (nums[middle] > target) {right = middle - 1; // target 在左区间，所以[left, middle - 1]} else if (nums[middle] < target) {left = middle + 1; // target 在右区间，所以[middle + 1, right]} else { // nums[middle] == targetreturn middle; // 数组中找到目标值，直接返回下标}}// 未找到目标值return -1;}};

• 左闭右开

class Solution {
public:int search(vector<int>& nums, int target) {int left = 0;int right = nums.size(); // 定义target在左闭右开的区间里，即：[left, right)while (left < right) { // 因为left == right的时候，在[left, right)是无效的空间，所以使用 <int middle = left + ((right - left) >> 1);if (nums[middle] > target) {right = middle; // target 在左区间，在[left, middle)中} else if (nums[middle] < target) {left = middle + 1; // target 在右区间，在[middle + 1, right)中} else { // nums[middle] == targetreturn middle; // 数组中找到目标值，直接返回下标}}// 未找到目标值return -1;}
};

//左闭右闭
public class Solution {  public int Search(int[] nums, int target) {int left = 0;int right = nums.Length - 1;while(left <= right){int mid = (right - left ) / 2 + left;if(nums[mid] == target){return mid;}else if(nums[mid] < target){left = mid+1;}else if(nums[mid] > target){right = mid-1;}}return -1;}
}//左闭右开
public class Solution{public int Search(int[] nums, int target){int left = 0;int right = nums.Length;while(left < right){int mid = (right - left) / 2 + left;if(nums[mid] == target){return mid;}else if(nums[mid] < target){left = mid + 1;}else if(nums[mid] > target){right = mid;}}return -1;}
}

27. 移除元素

// 时间复杂度：O(n)
// 空间复杂度：O(1)
class Solution {
public:int removeElement(vector<int>& nums, int val) {int slowIndex = 0;for (int fastIndex = 0; fastIndex < nums.size(); fastIndex++) {if (val != nums[fastIndex]) {nums[slowIndex++] = nums[fastIndex];}}return slowIndex;}
};

977. 有序数组的平方

public class Solution {public int[] SortedSquares(int[] nums) {int k = nums.Length - 1;int[] result = new int[nums.Length];for (int i = 0, j = nums.Length - 1;i <= j;){if (nums[i] * nums[i] < nums[j] * nums[j]) {result[k--] = nums[j] * nums[j];j--;} else {result[k--] = nums[i] * nums[i];i++;}}return result;}
}