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. Example
If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2. Challenge
If the count of numbers is bigger than MAXINT, can your code work properly?
跟Leetcode里search for a range挺像的，就是找到一个target之后，还要继续找它的左边沿。最后l指针超过r指针之后， l 指针会停在左边沿上
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, r = nums.length - 1;
int m = 0;
while (l <= r) {
m = (l + r) / 2;
if (nums[m] == target) break;
else if (nums[m] > target) r = m - 1;
else l = m + 1;
}
if (nums[m] != target) return -1;
l = 0;
r = m;
while (l <= r) {
m = (l + r) / 2;
if (nums[m] == target) {
r = m - 1;
}
else l = m + 1;
}
return l;
}
}