Binary Search (二分查找)

 

思路就是砍一半, 适用于sorted array. 时间复杂度O(lgn).

每次都是取中间的跟target比较, 比target小的话目标值肯定在lower和mid之间, 比target大的话在mid和high之间~

 

public int binarySearch(int num[], int target){
    int lower = 0;
    int high = num.length - 1;
    while(lower <= high){
        int mid = lower + (high - lower) / 2;
        if(num[mid] < target){
            lower = mid + 1;
        }else if(num[mid] > target)
            high = mid - 1;
        else
            return mid;
    }
    return lower;
}