搞懂二分查找

标签：static   else   arch   循环   bin   while   一个   退出   return   

class Solution {
    public static void main(String[] args) {
        int[] arr = new int[]{0, 1, 2, 2, 2, 3, 4, 5};
        //int index = binarySearch(arr, 2);
        //int index2 = leftBound(arr, 2);
        int index3 = rightBound(arr, 2);
        //System.out.println(index + ":" + arr[index]);
        //System.out.println(index2 + ":" + arr[index2]);
        System.out.println(index3 + ":" + arr[index3]);
    }

    public static int binarySearch(int[] nums, int target) {
        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 if (nums[mid] > target) {
                right = mid - 1;
            }
        }
        return -1;
    }
    
    /*
    * 找左边界和找右边界都有一个退出while循环的情况，即[left==right+1,right]
    * 所以:
    * 如果找左边界，返回left
    * 如果找右边界，返回right
    * 当然：
    * 返回之前，需要判断left或者right是否越界，即：left>=num.length和right<0
    * 并且判断nums[left/right] 是否等于 target 
    * */
    
    public static int leftBound(int[] nums, int target) {
        int left = 0;
        int 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 if (nums[mid] > target) {
                right = mid - 1;
            }
        }
        if (left >= nums.length || nums[left] != target)
            return -1;
        return left;
    }

    public static int rightBound(int[] nums, int target) {
        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 if (nums[mid] < target) {
                left = mid + 1;
            } else if (nums[mid] > target) {
                right = mid - 1;
            }
        }
        if (right < 0 || nums[right] != target)
            return -1;
        return right;
    }

}

搞懂二分查找

标签：static   else   arch   循环   bin   while   一个   退出   return   

原文地址：https://www.cnblogs.com/zyk98/p/14594788.html