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)/2
              if (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] == target
                  return 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] == target
                return 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;
    }
}