Search for a Range & Search Insert Position leetcode

二分查找算法应用

最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：https://segmentfault.com/a/1190000004288565

文章标签：

#数据结构与算法 #runtime

本文介绍了二分查找算法在寻找目标值范围及插入位置问题上的应用。通过两个具体实例展示了如何利用二分查找高效解决问题，并提供了简洁易懂的代码实现。

Search for a Range

Given a sorted array of integers, find the starting and ending
position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example, Given [5, 7, 7, 8, 8, 10] and target value 8, return [3,
4].

二分法

说明

用两次二分查找分别找出左右边界

复杂度

时间 O(logN) 空间 O(1)

代码

public int[] searchRange(int[] nums, int target) {
    int[] res = {-1, -1};
    if (nums == null || nums.length == 0) {
        return res;
    }
    int left = 0;
    int right = nums.length - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] < target) {
            left = mid + 1;
        } else {
            right = mid - 1;
        }
    }
    int start = 0;
    int end = nums.length - 1;
    while(start <= end) {
        int mid = start + (end - start) / 2;
        if (nums[mid] > target) {
            end = mid - 1;
        } else {
            start = mid + 1;
        }
    }
    if (left <= end) {
        res[0] = left;
        res[1] = end;
    }
    return res;
}

Search Insert Position

Given a sorted array and a target value, return the index if the
target is found. If not, return the index where it would be if it were
inserted in order.

You may assume no duplicates in the array.

Here are few examples. [1,3,5,6], 5 → 2 [1,3,5,6], 2 → 1 [1,3,5,6], 7
→ 4 [1,3,5,6], 0 → 0

二分法

复杂度

时间O(logn)，空间复杂度O(1)

代码

public int searchInsert(int[] nums, int target) {
    if (nums == null || nums.length == 0) {
        return 0;
    }
    int left = 0, right = nums.length - 1;
    while (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 left;
}