day1

1. [0704二分查找]

   class Solution {
       public int search(int[] nums, int target) {
           for (int i = 0; i <= nums.length-1; i++){
               if (nums[i] == target)
                   return i;
           }
           return -1;
       }
   }
   

   • 注意数组下标从0开始；没有考虑到本题数组元素有序性、唯一性，尤其是唯一性》》》》随想录说要用二分查找，我就写了下面代码：

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

   • 其实一开始先想用的是for循环，每次都很少想到while，二者还是有区别的；而且还遇到超出时间限制的错误，发现其实就是在while 循环中少了一句对mid重新赋值的语句；但结果还是不对那是因为没有注意到数组下表默认是从0开始，也就是说如果要返回数组第一个元素下标的话，那么返回值是0而不是1；最后更改如下：

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

   • 根据代码随想录把mid赋值语句合并到while开头这一处，另外在一开始就判断一下target值是否本来就不在范围内，再完善一下：

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

2. [027移除元素]

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

   • 暴力解法，我还给遇到问题了。。。原因是我变量设置太多了绕太多弯了，numLength和变量j好好看看，尤其是控制变量j的前面不就是j-1用上了么。但是没考虑到双循环可通过快慢指针编程单循环降低时间复杂度。明天优化
     ：）qly