# JAVA once clear binary search

• My code may not be very short, but I thought it's clear and easy to understand. The most important fact to realize is that no matter what, at least one half (either low ~mid, mid ~ high) is well sorted. We can use this to solve the problem case by case

``````    public int search(int[] nums, int target) {
int low = 0;
int high = nums.length - 1;
int mid = low + (high - low)/2;

while(low <= high){
int midVal = nums[mid];
if(target == midVal){
return mid;
}
// whole list is well sorted
if(midVal < nums[high] && midVal > nums[low]){
if(target < midVal){
high = mid - 1;
}
else{
low = mid + 1;
}
}
// right side is sorted
else if(midVal < nums[low]){
if(target < midVal){
high = mid - 1;
}
else{
if(nums[high] >= target){
low = mid + 1;
}
else{
high = mid - 1;
}
}
}
// left side is sorted
else{
if(target < midVal){
if(nums[low] <= target){
high = mid - 1;
}
else{
low = mid + 1;
}
}
else{
low = mid + 1;
}
}
mid = low + (high - low)/2;
}

return -1;
}
``````

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