89 lines
2.8 KiB
Java
89 lines
2.8 KiB
Java
package binary_search;
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/**
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* 题目： 34. 在排序数组中查找元素的第一个和最后一个位置 (searchRange)
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* 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。
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* 如果数组中不存在目标值 target，返回 [-1, -1]。
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* 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。
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* 示例 1：
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输入：nums = [5,7,7,8,8,10], target = 8
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输出：[3,4]
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* 链接：https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/
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*/
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public class SearchRange {
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//极端情况下不是O(log n)
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public int[] searchRange(int[] nums, int target) {
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int left=0,right=nums.length-1;
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int tpleft=-1,tpright=-1;
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while (left<=right){
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int mid=(left+right)/2;
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if(target==nums[mid]){
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tpleft=mid-1;
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tpright=mid+1;
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while (tpleft>=left) {
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if (target == nums[tpleft])
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tpleft--;
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else
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break;
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}
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tpleft++;
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while (tpright<=right) {
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if (target == nums[tpright])
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tpright++;
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else
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break;
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}
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tpright--;
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break;
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}
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else if(target>nums[mid])
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left=mid+1;
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else
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right=mid-1;
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}
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return new int[]{tpleft,tpright};
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}
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private int findLeft(int[] nums, int target) {
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int left = 0, right = nums.length - 1;
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int index = -1;
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (nums[mid] >= target) {
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right = mid - 1;
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} else {
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left = mid + 1;
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}
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if (nums[mid] == target) {
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index = mid;
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}
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}
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return index;
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}
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// 找到目标值的最后一次出现位置
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private int findRight(int[] nums, int target) {
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int left = 0, right = nums.length - 1;
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int index = -1;
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while (left <= right) {
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int mid = left + (right - left) / 2;
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if (nums[mid] <= target) {
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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if (nums[mid] == target) {
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index = mid;
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}
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}
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return index;
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}
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public int[] searchRange1(int[] nums, int target) {
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int leftIndex = findLeft(nums, target);
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int rightIndex = findRight(nums, target);
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return new int[]{leftIndex, rightIndex};
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}
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}
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