Binary Search

For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Example

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

这是一个经典的binary serch的模板

 1.start+1 < end

2. mid = start + (end-start/)2

3. nums[mid] <, ==,> target 的三种情况

4. 是return start end 还是-1

class Solution {
    /**
     * @param nums: The integer array.
     * @param target: Target to find.
     * @return: The first position of target. Position starts from 0.
     */
    public int binarySearch(int[] nums, int target) {
        if (nums.length == 0){
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        int mid;
        while (start + 1 < end){
            mid = start + (end - start) / 2;
            if (target > nums[mid]){
                start = mid;
            } else if (target < nums[mid]) {
                end = mid;
            } else {
                end = mid;
            }
        }
        if (nums[start] == target){
            return start;
        } else if (nums[end] == target){
            return end;
        } else {
            return -1;
        }
    }
}