力扣labuladong一刷day22天二分搜索共2题
一、704. 二分查找
题目链接:https://leetcode.cn/problems/binary-search/
思路:典型的二分查找,如果是左闭右闭那么说明left <= right 。如果左闭右开那么说明 left < right
class Solution {
public int 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) {
return mid;
}else if (nums[mid] < target) {
left = mid + 1;
}else {
right = mid - 1;
}
}
return -1;
}
}
二、34. 在排序数组中查找元素的第一个和最后一个位置
题目链接:https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/
思路:分别寻找左右边界。
class Solution {
public int[] searchRange(int[] nums, int target) {
int left = findLeft(nums, target);
int right = findRight(nums, target);
return new int[]{left, right};
}
int findLeft(int[] nums, int target) {
int left = 0, right = nums.length-1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] >= target) {
right = mid-1;
}else {
left = mid+1;
}
}
if (left < 0 || left >= nums.length) return -1;
return nums[left] == target ? left : -1;
}
int findRight(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 {
right = mid-1;
}
}
if (right < 0 || right >= nums.length) return -1;
return nums[right] == target ? right : -1;
}
}