二分查找

题目链接

https://programmercarl.com/0704.%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE.html#_704-%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE

思路

二分查找关键在于边界如何处理，有左闭右开和左闭右闭两种形式

左闭右开

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

左闭右闭

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

移除元素

题目链接

https://programmercarl.com/0027.%E7%A7%BB%E9%99%A4%E5%85%83%E7%B4%A0.html

思路

暴力解法，从头到尾遍历，遇到要删除的元素，后面元素左移覆盖

public int removeElement(int[] nums, int val) {
      int size = nums.length;
      for(int i = 0; i < size; i++) {
          if (nums[i] == val) {
              for(int j = i + 1; j < size; j++) {
                   nums[j - 1] = nums[j];
              }
             size--;
             i--;
          }
      }
      return size;
}

快慢指针法
慢指针指向要删除的元素，快指针往前走，直到找到符合条件的，和慢指针交换，慢指针变为快指针

  public int removeElement(int[] nums, int val) {
    int left = 0, right = 0;          
    while(right < nums.length) {
        if (nums[left] == val) {
            while(right < nums.length && nums[right] == val) {
                right++;
            }
            if (right < nums.length) {
                nums[left++] = nums[right++];
                nums[right - 1] = val;
            }
            
        } else {
            left++;
            right++;
        }
     }
    return left;
}

• 没有理解快慢指针精髓
  优先移动快指针，而且移动的是快指针，那就应该用快指针比较，不能用慢指针
  参考卡哥的方法

    public int removeElement(int[] nums, int val) {
    int left = 0, right = 0;          
    while(right < nums.length) {
        if (nums[right] != val) {
            nums[left++] = nums[right++];
            
        } else {
            right++;
        }
     }
    return left;
  }
  

还有双指针写法，从左找=指定值元素，从右找<>指定值元素，交换

    public int removeElement(int[] nums, int val) {
         int left = 0,right = nums.length - 1;
          while(left <=right) {
            while(left <= right && nums[left] != val) {
                     left++;
            }
            while(left <= right && nums[right] == val) {
                    right--;
            }
            if (left < right) {
                   nums[left++] = nums[right--];
           }
         }
         return left;
    }