Binary Search

/相邻就退出

//start = 1, end = 2 就退出
//int mid = (start + end) /2
//start, end ~ 2^31
//Consider special case: [2,2], target = 2
//mid = (0 + 1) / 2 = 0;
// start < end
//dead loop
while (start + 1 < end) {
int mid = (end - start) /2 + start;
if (target == nums[mid]) {
end = mid;
} else if (target < nums[mid]) {
} else { start = mid}
//first position, 先看 start, if last position, 先看end
if (nums[start] = target) {
return start;
}
if (nums[end] == target) {
return end;
}
//start ..............mid..............end
//               target

class Solution {
public:
    /**
     * @param nums: The integer array.
     * @param target: Target number to find.
     * @return: The first position of target. Position starts from 0. 
     */
    int binarySearch(vector<int> &array, int target) {
        // write your code here
        if (array.size() == 0) {
            return -1;
        }
        int start = 0;
        int end = array.size() - 1;
        int mid;

        while (start + 1 < end) {
            mid = (start + end) / 2;
            if (array[mid] < target) {
                start = mid;
            } 
            else if (array[mid] > target) {
                end = mid;
            } else if (array[mid] == target) {
                end = mid;
            }
        }

        if (array[start] == target) {
            return start;
        }
        if (array[end] == target) {
            return end;
        }

        return -1;
    }
};