Binary Search Common Problems

find the first number index that is larger or equals to target.
(就是说返回target的最后一个index,或者比target大的第一个index,如果一定要确定放在target的最后一个index上，那最后一定要check right指针是不是对应target)

**//if we only have one element, and it's keeping >
int binarySearch(int[] arr, int target){
    int left = 0, right = arr.length - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] < target) {
            left = mid + 1;
        }else{ //if arr[mid] >= target
            right = mid;
        }
    }
    return right;
}** //not sure if this code is right
int binarySearch(int[] arr, int target){
    int left = 0, right = arr.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] < target) {
            left = mid + 1;
        }else{ //if arr[mid] >= target
            right = mid - 1;
        }
    }
    if (right == arr.length - 1) {
        return -1;
    }
    return right + 1;
}

find the first index that is larger than target:

int binarySearch(int[] arr, int target){
    int left = 0, right = arr.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] <= target) {
            left = mid + 1;
        }else{ //if arr[mid] > target
            right = mid;
        }
    }
    return right;
}

find the last index that is less or equals to target:

find the last index that is less than target:(?)

int binarySearch(int[] arr, int target){
    int left = 0, right = arr.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] < target) {
            left = mid + 1;
        }else{ //if arr[mid] >= target
            right = mid;
        }
    }
    return right - 1;;
}

find the first index that is equals to target:

int binarySearch(int[] arr, int target){
    int left = 0, right = arr.length - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] < target) {
            left = mid + 1;
        }else{ //if arr[mid] >= target
            right = mid;
        }
    }
    if (arr[left] == val) {
        return left;
    }
    return -1;;
}

find the last index that equals to target:

int binarySearch(int[] arr, int target){
    int left = 0, right = arr.length - 1;
    while (left < right) {
        int mid = left + (right - left) / 2;
        if (arr[mid] <= target) {
            left = mid;
        }else{ //if arr[mid] >= target
            right = mid - 1;;
        }
    }
    if (arr[right] == val) {
        return right;
    }
    return -1;;
}

find the index range of given target:

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if(nums == null || nums.length == 0) return new int[]{-1,-1};
        
        //int[] res = new int[];
        
        int left = 0;
        int right = nums.length - 1;
        while(left <= right) {
            int mid = (left + right) / 2;
            if(nums[mid] == target) {
                int i = mid;
                int j = mid;
                while(i >= 0 && nums[i] == target) {
                    i--;
                }
                while(j <= nums.length -1 && nums[j] == target) {
                    j++;
                }
                return new int[]{i+1, j-1};
            }
            if(nums[mid] > target) {
                right = mid - 1;
            }
            if(nums[mid] < target) {
                left = mid + 1;
            }
        }
        return new int[]{-1,-1};
    }
}