二分查找--簡潔、安全的實現

二分適用於：數據呈現某種規律（有序） ;
算法 時間複雜度 log(n);

實現

1. 循環實現，沒瑕疵的

  public static int binarySearch(int[] arr, int num) {
        int nuMax = arr.length - 1;
        int nuMin = 0;
        while (nuMin <= nuMax) {
            //防止溢出
            int middle = ((nuMax - nuMin) >> 1) + nuMin;
//            int middle = (nuMax+nuMin)>>1 + 1;
            if (arr[middle] == num) {
                return middle;
            } else if (arr[middle] < num) {
                nuMin = middle + 1;
            } else{
                nuMax = middle - 1;
            }
        }
        return -1;
    }

2. 遞歸

    public static int binarySearch(int[] arr, int num, int nuMin, int nuMax) {
        int middle = ((nuMax - nuMin) >> 1) + nuMin;
        while (nuMax >= nuMin) {
            if (arr[middle] == num)
                return middle;
            else if (arr[middle]< num){
               return  binarySearch(arr,num,middle+1,nuMax);
            }else
                return binarySearch(arr,num,nuMin,middle -1);
        }
        return -1;
    }