模板一：
public static int erfenchazhao(int[] nums,int target){
if(nums==null||nums.length==0)return -1;
int left=0,right=nums.length-1;
whilt(left<=right){
int mid=left+(right-left)/2;
if(nums[mid]==target)return mid;
else if(nums[mid]<target)left=mid+1;
else right=mid-1;
}
return -1;//left>right
}
模板二
int binarySearch(int[] nums, int target){
if(nums == null || nums.length == 0)
return -1;
int left = 0, right = nums.length;
while(left < right){
// Prevent (left + right) overflow
int mid = left + (right - left) / 2;
if(nums[mid] == target){ return mid; }
else if(nums[mid] < target) { left = mid + 1; }
else { right = mid; }
}
// Post-processing:
// End Condition: left == right
if(left != nums.length && nums[left] == target) return left;
return -1;
}
模板三
int binarySearch(int[] nums, int target) {
if (nums == null || nums.length == 0)
return -1;
int binarySearch(int[] nums, int target) {
if (nums == null || nums.length == 0)
return -1;
int left = 0, right = nums.length - 1;
while (left + 1 < right){
// Prevent (left + right) overflow
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] < target) {
left = mid;
} else {
right = mid;
}
}
// Post-processing:
// End Condition: left + 1 == right
if(nums[left] == target) return left;
if(nums[right] == target) return right;
return -1;
}
