import java.util.Arrays;
// 1235. Maximum Profit in Job Scheduling
class Solution {
public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
int n = startTime.length;
int[][] jobs = new int[n + 1][];
jobs[0] = new int[]{0, 0, 0};
for (int i = 0; i < n; ++i) {
jobs[i + 1] = new int[]{startTime[i], endTime[i], profit[i]};
}
Arrays.sort(jobs, (j1, j2) -> j1[1] - j2[1]);
int[] dp = new int[n + 1];
for (int i = 1; i <= n; ++i) {
int j = binarySearch(i, jobs[i][0], jobs);
dp[i] = Math.max(dp[i - 1], dp[j] + jobs[i][2]);
}
return dp[n];
}
int binarySearch(int rightest, int latestEndTime, int[][] jobs) {
int left = 0;
int right = rightest;
while (left < right) {
int mid = left + (right - left) / 2;
if (jobs[mid][1] <= latestEndTime) {
left = mid + 1;
} else {
right = mid;
}
}
return left - 1;
}
}
学习笔记：
今天是一道困难题，需要用到动态规划 + 二分搜素。
今天做到二分搜索的时候遇到了疑惑，刚好今天六点多就结束加班了，就学习了一波。
二分搜索分为三种，搜索一个数找不到就返回-1的，搜索左边界的，搜索右边界的。
先看循环条件：
搜索一个数的话，left <= right
搜索边界的话，left < right
核心情况在于nums[mid] == target
搜索一个数的话，返回mid
搜索左边界，right = mid
搜索右边界，left = mid + 1
返回值要注意的就是
搜索左边界，返回left或right
搜索右边界，返回left - 1或right - 1
// 搜索一个数
int binarySearch(int[] nums, int target) {
int left = 0;
int right = nums.length - 1; // 注意
while(left <= right) {
int mid = (right + left) / 2;
if(nums[mid] == target)
return mid;
else if (nums[mid] < target)
left = mid + 1; // 注意
else if (nums[mid] > target)
right = mid - 1; // 注意
}
return -1;
}
// 搜索左边界
int left_bound(int[] nums, int target) {
if (nums.length == 0) return -1;
int left = 0;
int right = nums.length; // 注意
while (left < right) { // 注意
int mid = (left + right) / 2;
if (nums[mid] == target) {
right = mid;
} else if (nums[mid] < target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid; // 注意
}
}
return left;
}
// 搜索右边界
int right_bound(int[] nums, int target) {
if (nums.length == 0) return -1;
int left = 0, right = nums.length;
while (left < right) {
int mid = (left + right) / 2;
if (nums[mid] == target) {
left = mid + 1; // 注意
} else if (nums[mid] < target) {
left = mid + 1;
} else if (nums[mid] > target) {
right = mid;
}
}
return left - 1; // 注意
}
