Binary search is a famous question in algorithm.
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.
样例
If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.
挑战
If the count of numbers is bigger than MAXINT, can your code work properly?
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) {
int l = 0;
int h = nums.length - 1;
while (l <= h) {
int m = l / 2 + h / 2 + (l % 2 & h % 2);
if (nums[m] == target) {
while (m > 0 && nums[m - 1] == target)
m--;
return m;
} else if (nums[m] > target) {
h = m - 1;
} else if (nums[m] < target) {
l = m + 1;
}
}
return -1;
}
}