解题报告 - 在排序元素中查找元素的第一个和最后一个位置

解题报告 - 在排序元素中查找元素的第一个和最后一个位置

LeetCode 在排序元素中查找元素的第一个和最后一个位置

@TOC

题目描述

给你一个按照非递减顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。

如果数组中不存在目标值 target，返回 [-1, -1]。

你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。

示例：

<pre class="md-fences md-end-block ty-contain-cm modeLoaded" spellcheck="false" lang="" cid="n11" mdtype="fences" style="box-sizing: border-box; overflow: visible; font-family: var(--monospace); font-size: 0.9em; display: block; break-inside: avoid; text-align: left; white-space: normal; background-image: inherit; background-size: inherit; background-attachment: inherit; background-origin: inherit; background-clip: inherit; background-color: rgb(248, 248, 248); position: relative !important; border: 1px solid rgb(231, 234, 237); border-top-left-radius: 3px; border-top-right-radius: 3px; border-bottom-right-radius: 3px; border-bottom-left-radius: 3px; padding: 8px 4px 6px; margin-bottom: 15px; margin-top: 15px; width: inherit; caret-color: rgb(51, 51, 51); color: rgb(51, 51, 51); font-style: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: auto; text-indent: 0px; text-transform: none; widows: auto; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; text-decoration: none; background-position: inherit; background-repeat: inherit;">输入：nums = [5,7,7,8,8,10], target = 8 输出：[3,4]</pre>

提示：

<pre class="md-fences md-end-block ty-contain-cm modeLoaded" spellcheck="false" lang="" cid="n13" mdtype="fences" style="box-sizing: border-box; overflow: visible; font-family: var(--monospace); font-size: 0.9em; display: block; break-inside: avoid; text-align: left; white-space: normal; background-image: inherit; background-size: inherit; background-attachment: inherit; background-origin: inherit; background-clip: inherit; background-color: rgb(248, 248, 248); position: relative !important; border: 1px solid rgb(231, 234, 237); border-top-left-radius: 3px; border-top-right-radius: 3px; border-bottom-right-radius: 3px; border-bottom-left-radius: 3px; padding: 8px 4px 6px; margin-bottom: 15px; margin-top: 15px; width: inherit; caret-color: rgb(51, 51, 51); color: rgb(51, 51, 51); font-style: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: auto; text-indent: 0px; text-transform: none; widows: auto; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; text-decoration: none; background-position: inherit; background-repeat: inherit;">0 <= nums.length <= 105
-109 <= nums[i] <= 109
nums 是一个非递减数组
-109 <= target <= 109</pre>

一、解题关键词

<pre class="md-fences md-end-block ty-contain-cm modeLoaded" spellcheck="false" lang="" cid="n15" mdtype="fences" style="box-sizing: border-box; overflow: visible; font-family: var(--monospace); font-size: 0.9em; display: block; break-inside: avoid; text-align: left; white-space: normal; background-image: inherit; background-size: inherit; background-attachment: inherit; background-origin: inherit; background-clip: inherit; background-color: rgb(248, 248, 248); position: relative !important; border: 1px solid rgb(231, 234, 237); border-top-left-radius: 3px; border-top-right-radius: 3px; border-bottom-right-radius: 3px; border-bottom-left-radius: 3px; padding: 8px 4px 6px; margin-bottom: 15px; margin-top: 15px; width: inherit; caret-color: rgb(51, 51, 51); color: rgb(51, 51, 51); font-style: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: auto; text-indent: 0px; text-transform: none; widows: auto; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; text-decoration: none; background-position: inherit; background-repeat: inherit;">非递减
复杂度为O（logn）</pre>

二、解题报告

1.思路分析

1. 根据时间复杂度要求 可推断 需要二分思想解决问题

2. 需要找到两个坐标 left right ，分治思想，先找到一个坐标

3. Boolean flag 用来标识 用来区分左边界寻找，还是右边界寻找

2.时间复杂度

3.代码示例

<pre class="md-fences md-end-block ty-contain-cm modeLoaded" spellcheck="false" lang="java" cid="n27" mdtype="fences" style="box-sizing: border-box; overflow: visible; font-family: var(--monospace); font-size: 0.9em; display: block; break-inside: avoid; text-align: left; white-space: normal; background-image: inherit; background-size: inherit; background-attachment: inherit; background-origin: inherit; background-clip: inherit; background-color: rgb(248, 248, 248); position: relative !important; border: 1px solid rgb(231, 234, 237); border-top-left-radius: 3px; border-top-right-radius: 3px; border-bottom-right-radius: 3px; border-bottom-left-radius: 3px; padding: 8px 4px 6px; margin-bottom: 15px; margin-top: 15px; width: inherit; caret-color: rgb(51, 51, 51); color: rgb(51, 51, 51); font-style: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: auto; text-indent: 0px; text-transform: none; widows: auto; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; text-decoration: none; background-position: inherit; background-repeat: inherit;">class Solution {
//非递减顺序 递增 或者相等
public int[] searchRange(int[] nums, int target) {
if (null == nums || nums.length < 1) return new int[]{-1, -1};
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};
}

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

}
return ans;
}
}</pre>

4.知识点

<pre class="md-fences md-end-block ty-contain-cm modeLoaded" spellcheck="false" lang="" cid="n29" mdtype="fences" style="box-sizing: border-box; overflow: visible; font-family: var(--monospace); font-size: 0.9em; display: block; break-inside: avoid; text-align: left; white-space: normal; background-image: inherit; background-size: inherit; background-attachment: inherit; background-origin: inherit; background-clip: inherit; background-color: rgb(248, 248, 248); position: relative !important; border: 1px solid rgb(231, 234, 237); border-top-left-radius: 3px; border-top-right-radius: 3px; border-bottom-right-radius: 3px; border-bottom-left-radius: 3px; padding: 8px 4px 6px; margin-bottom: 15px; margin-top: 15px; width: inherit; caret-color: rgb(51, 51, 51); color: rgb(51, 51, 51); font-style: normal; font-variant-caps: normal; font-weight: normal; letter-spacing: normal; orphans: auto; text-indent: 0px; text-transform: none; widows: auto; word-spacing: 0px; -webkit-text-size-adjust: auto; -webkit-text-stroke-width: 0px; text-decoration: none; background-position: inherit; background-repeat: inherit;"></pre>

三、总结

1. 目的是为了找到一个区间 ==target

2. 找到左左边后，要想办法找到右坐标 且 左右坐标要有区别

3. 即，左边坐标一定要让right --

4. 右边坐标 == target的时候 想办法让left ++

5. 两个的区别就在于 mid >= target ? 大于 则右坐标-- 得到右边界：左坐标++ 找到左边界