LintCode: 二分查找

target

O(logn)

-1

[1, 2, 3, 3, 4, 5, 10]

3

2

public int binarySearch(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}

// write your code here
int pre = 0, end = nums.length-1;
while(pre<end){
int middle = (pre+end)/2;
if(nums[middle]<target){
pre = middle+1;
}else if(nums[middle]>target){
end = middle-1;
}else{
end = middle;  // 这一步很重要，需要求的是第一次出现的下标
}
}
if(nums[pre]==target){
return pre;
}

return -1;
}