查找算法---二分查找

二分查找

二分查找的思想:利用已排序的数组，每次都对半的减少查找的数量，直到找到对应的数据，时间复杂度为O(logn),比较适用于数据量适中的，因为局限于已排序的，所以对于无序的都需要进行一次排序操作。

// 递归版本

private int binarySearch2(int[] arr,int result){
        return binarySearch2(arr,0,arr.length - 1,result);
    }

    private int binarySearch2(int[] arr, int low, int high, int result) {
        if (low > high) return -1;
        //使用位运算
        int mid =  low + ((high - low)  >> 1);
        if (arr[mid] == result){
            return result;
        }else if (arr[mid] > result){
            return binarySearch2(arr,low,mid - 1,result);
        }else{
            return binarySearch2(arr,mid + 1,high,result);
        }
    }

// 非递归版本
private int binarySearch(int[] arr,int result){
        if (arr.length == 0) {
            return -1;
        }
        Arrays.sort(arr);
        int low = 0;
        int high = arr.length - 1;
        while (low <= high){
            // 不建议使用这样 建议使用 low + ((high - low) >> 1)
            int mid = (low + high) / 2;
            if (arr[mid] > result){
                high = mid - 1;
            }else if (arr[mid] < result){
                low = mid + 1;
            }else{
                return mid;
            }
        }
        return -1;
    }

二分查找变形题

 /**
     * 找到第一个出现的数字
     * @param sortArr
     * @param num
     * @return
     */
    private int findFirstNum(int[] sortArr,int num){
        int low = 0;
        int high = sortArr.length - 1;
        while (low <= high){
            int mid = low + ((high - low) >> 1);
            if (sortArr[mid] > num ){
                high = mid + 1;
            }else if (sortArr[mid] < num){
                low = mid - 1;
            }else{
                // 找到对应的 但是此时不确定是不是最先出现的
                if (mid == 0 || sortArr[mid - 1] != num){
                    return mid;
                }else{
                    high = mid - 1;
                }
            }
        }
        return -1;
    }

    /**
     * 查找最后一个
     * @param arr
     * @param num
     * @return
     */
    private int findLastNum(int[] arr,int num){
        int low = 0;
        int high = arr.length - 1;
        while (low <= high){
            int mid = low + ((high - low) >> 1);
            if (arr[mid] > num ){
                high = mid + 1;
            }else if (arr[mid] < num){
                low = mid - 1;
            }else{
                // 找到对应的 但是此时不确定是不是最先出现的
                if (mid == arr.length - 1 || arr[mid + 1] != num){
                    return mid;
                }else{
                   low = mid + 1;
                }
            }
        }
        return -1;
    }

    private int findFirstBigOne(int[] arr,int num){
        int low = 0;
        int high = arr.length - 1;
        while (low <= high){
            int mid = low + ((high - low) >> 1);
            if (arr[mid] >= num){
                if (mid == 0 || arr[mid - 1] < num){
                    return mid;
                }else{
                    high = mid - 1;
                }
            } else{
                low = mid + 1;
            }
        }
        return -1;
    }

    private int findLastSmallOne(int[] arr,int num){
        int low = 0;
        int high = arr.length - 1;
        while (low <= high){
            int mid = low + ((high - low) >> 1);
            if (arr[mid] < num){
                high = mid - 1;
            } else{
                if (mid == arr.length - 1 || arr[mid + 1] > num){
                    return mid;
                }else{
                    low = mid + 1;
                }

            }
        }
        return -1;
    }