二分法板子

二分法常常使用，根据统计只用10%的程序员才可以完美写出二分，二分的关键在于对条件、边界的判定，而下面这个模板是一个万能模板，转自知乎。

模板如下，

public int[] searchRange(int[] nums, int target) {
            if (nums ==null || nums.length == 0)
                return new int[]{-1,-1};
    
    
            int[] result = new int[2];
    
            int left = 0;
            int right = nums.length - 1;
            while (left < right){
                int mid = left + (right -left) / 2;
                if (nums[mid]  < target) left = mid + 1;
                else right = mid;
            }
            if (nums[left] != target)
            return -1;
}

注意的是left最后要进行检验，得到的是第一个target的值。

至于如何证明这个是一个完美模板还需要看原地址的证明。

源地址：https://www.zhihu.com/question/36132386/answer/530313852

博主只是使用java代码重构了一遍。

最后博主使用了这个板子写了两道关于二分的leetcode的题。

第一题 Find First and Last Position of Element in Sorted Array
https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/

  public int[] searchRange(int[] nums, int target) {
                if (nums ==null || nums.length == 0)
                    return new int[]{-1,-1};


                int[] result = new int[2];

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


            int index1 = left;
            for (int i = left; i < nums.length; i++){
                if (target == nums[i]) index1 = i;
                else break;
            }

            int index2 = left;
            for (int i = left; i >= 0; i--){
                if (target == nums[i]) index2 = i;
                else break;
            }
            result [0] = index2; result[1] = index1;
            return result;
        }

第二题 Search in Rotated Sorted Array

public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0)
            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;

            if (nums[mid] < nums[right] ){
                if (nums[mid] < target &&  nums[right] >=target) left = mid + 1;
                else right = mid;
            }else if (nums[mid] > nums[right]){
                if (nums[left] <= target && nums[mid] >target) right = mid ;
                else left =mid + 1;
            }

        }

        if(nums[left] == target)
            return left;
        else
        return -1;
    }