Java二分查找左右边界

有序数组左右全闭的写法：

查找左边界例子:

1 2 2 3 4
0 1 2 3 4

如果查找的元素是2 在数组中存在
mid = 2 target = a[mid]
left = 0
right = 4

mid = 2 target = a[mid]
left = 0
right = 1

mid = 0  a[mid] < target
left = mid+1 = 1
right = 0

left >= right成立
return left=1;

如果查找的target=0 即小于数组中最小的元素
那么
mid = 2  a[mid] > target
left = 0
right = 4

mid = 2  a[mid] > target
left = 0
right = 1

mid = 0  a[mid] > target
left = 0
right = 0

mid = 0  a[mid] > target
left = 0
right = -1

left >= right成立
a[left]!=target成立
return -1;

如果查找的target = 5 即大于数组中最大的元素
那么
mid = 2  a[mid] < target
left = 3
right = 4

mid = 3  a[mid] < target
left = 4
right = 4

mid = 4  a[mid] < target
left = 5
right = 4

left >= a.length成立
return -1;

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

int left_bound(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1;
        } else if (nums[mid] == target) {
            // 别返回，锁定左侧边界
            right = mid - 1;
        }
    }
    // 最后要检查 left 越界的情况
    if (left >= nums.length || nums[left] != target)
        return -1;
    return left;
}


int right_bound(int[] nums, int target) {
    int left = 0, right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] < target) {
            left = mid + 1;
        } else if (nums[mid] > target) {
            right = mid - 1;
        } else if (nums[mid] == target) {
            // 别返回，锁定右侧边界
            left = mid + 1;
        }
    }
    // 最后要检查 right 越界的情况
    if (right < 0 || nums[right] != target)
        return -1;
    return right;
}