leetcode每日一题-34:在排序数组中查找元素的第一个和最后一个位置

这个其实就是昨天题目的升级版本,掌握应用两种二分的模板即可,这两种模板可以应用在90%的整数二分问题当中

• c++

class Solution {
public:
    // 两个整数二分模板的应用
    vector searchRange(vector& nums, int target) {
        int n = nums.size();
        vector res;

        // 为空则一定无答案
        if(n == 0) return {-1, -1};

        // 找到第一个大于等于target的值
        int l = 0, r = n - 1;
        while(l < r)
        {
            int mid = l + r >> 1;
            if(nums[mid] >= target) r = mid;
            else l = mid + 1;
        }

        // 如果这个值等与target我们继续搜索taret的右边界
        if(nums[l] == target)
        {
            res.push_back(l);
            
            // 找到第一个小于等于target的值,注意mid值的变化
            l = 0, r = n-1;
            while(l < r)
            {
                int mid = l + r + 1 >> 1;
                if(nums[mid] <= target) l = mid;
                else r = mid - 1;
            }
            // cout << l << endl;
            res.push_back(l);
            return res;
        }
        else return {-1, -1};
    }
};

• java

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int leftIdx = binarySearch(nums, target, true);
        int rightIdx = binarySearch(nums, target, false) - 1;
        if (leftIdx <= rightIdx && rightIdx < nums.length && nums[leftIdx] == target && nums[rightIdx] == target) {
            return new int[]{leftIdx, rightIdx};
        } 
        return new int[]{-1, -1};
    }

    public int binarySearch(int[] nums, int target, boolean lower) {
        int left = 0, right = nums.length - 1, ans = nums.length;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] > target || (lower && nums[mid] >= target)) {
                right = mid - 1;
                ans = mid;
            } else {
                left = mid + 1;
            }
        }
        return ans;
    }
}

作者：LeetCode-Solution