• LeetCode第七天


    数组 Medium

    40.(162)Find Peak Element

    JAVA
    //斜率思想,二分法
    class Solution {
        public int findPeakElement(int[] nums) {
            int l=0,r=nums.length-1;
            while(l<r){      
                int mid = (r+l)/2;
                if(nums[mid]>nums[mid+1])
                    r = mid;
                else
                    l = mid+1;
            }
            return l;
        }
    }
    

    41.(731)My Calendar II

    JAVA
    public class MyCalendarTwo {
        List<int[]> calendar;
        List<int[]> overlaps;//已经重叠过一次的区间
    
        MyCalendarTwo() {
            calendar = new ArrayList();
            overlaps = new ArrayList();
        }
    
        public boolean book(int start, int end) {
            for (int[] iv: overlaps) {
                if (iv[0] < end && start < iv[1]) return false;
            }
            for (int[] iv: calendar) {
                if (iv[0] < end && start < iv[1])
                    overlaps.add(new int[]{Math.max(start, iv[0]), Math.min(end, iv[1])});
            }
            calendar.add(new int[]{start, end});
            return true;
        }
    }
    

    42.(153)Find Minimum in Rotated Sorted Array

    JAVA
    class Solution {
        public int findMin(int[] nums) {
            int l = 0;
            int r = nums.length-1;
            while(l<r){
                int mid = (l+r)/2;
                if(nums[mid]<nums[r])
                    r = mid;
                else
                    l = mid+1;
            }
            return nums[r];
        }
    }
    
    class Solution {
        public int findMin(int[] nums) {
           return find(nums,0,nums.length-1);
        }
        
        public int find(int[] nums, int l, int r) {
            if(nums[l] <= nums[r]) {
                return nums[l];
            }
            int mid = (l + r) / 2;
            return Math.min(find(nums,l,mid),find(nums,mid+1,r));
        }
    }
    

    43.(152)Maximum Product Subarray

    JAVA
    class Solution {
        public int maxProduct(int[] nums) {
            if(nums.length == 0){
                return 0;
            }
            int maxPre = nums[0];
            int minPre = nums[0];
            int max = nums[0];
            for(int i =1;i<nums.length;i++){
                int maxHere = Math.max(Math.max(maxPre*nums[i],minPre*nums[i]),nums[i]);
                int minHere = Math.min(Math.min(maxPre*nums[i],minPre*nums[i]),nums[i]);
                max = Math.max(max,maxHere);
                maxPre = maxHere;
                minPre = minHere;
            }
            return max;
        }
    }
    

    44.(611)Valid Triangle Number

    JAVA
    class Solution {
        public int triangleNumber(int[] nums) {
            int count = 0;
            Arrays.sort(nums);
            for(int i =0;i<nums.length-2;i++){
                if(nums[i]==0)
                    continue;
                int k = i+2;
                for(int j = i+1;j<nums.length-1;j++){
                    while(k<nums.length&&nums[i]+nums[j]>nums[k])
                        k++;
                    count += k-j-1;
                }
            }
            return count;
        }
    }
    

    45.(621)Task Scheduler

    JAVA
    //计算休眠时间,再加上任务时间等于总时间
    class Solution {
        public int leastInterval(char[] tasks, int n) {
            int[] map = new int[26];
            for(char c : tasks)
                map[c-'A']++;
            Arrays.sort(map);
            int idle = (map[25] -1)*n;
            for(int i=24;i>=0&&map[i]>0;i--){
                idle -= Math.min(map[i],map[25]-1);
            }
            return idle >0 ? tasks.length+idle:tasks.length;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/guoyaohua/p/8254275.html
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