128th LeetCode Weekly Contest Capacity To Ship Packages Within D Days

A conveyor belt has packages that must be shipped from one port to another within D days.

The i-th package on the conveyor belt has a weight of weights[i].  Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days.

 

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5 Output: 15 Explanation: A ship capacity of 15 is the minimum to ship all the packages in 5 days like this: 1st day: 1, 2, 3, 4, 5 2nd day: 6, 7 3rd day: 8 4th day: 9 5th day: 10 Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed. 

Example 2:

Input: weights = [3,2,2,4,1,4], D = 3 Output: 6 Explanation: A ship capacity of 6 is the minimum to ship all the packages in 3 days like this: 1st day: 3, 2 2nd day: 2, 4 3rd day: 1, 4 

Example 3:

Input: weights = [1,2,3,1,1], D = 4 Output: 3 Explanation: 1st day: 1 2nd day: 2 3rd day: 3 4th day: 1, 1 

 

Note:

1. 1 <= D <= weights.length <= 50000
2. 1 <= weights[i] <= 500

題意看錯了，應該是找最大值。這樣妥妥的二分啊

當sum>max時候，咱們開始二分。

瘋狂找（ai+....+ax） > mid 這個有幾段 getRequiredPainters這個就這麼用的

int getMax(int A[], int n) { int max = INT_MIN; for (int i = 0; i < n; i++) { if (A[i] > max) max = A[i]; } return max; } int getSum(int A[], int n) { int total = 0; for (int i = 0; i < n; i++) total += A[i]; return total; } int getRequiredPainters(int A[], int n, int maxLengthPerPainter) { int total = 0, numPainters = 1; for (int i = 0; i < n; i++) { total += A[i]; if (total > maxLengthPerPainter) { total = A[i]; numPainters++; } } return numPainters; } int BinarySearch(int A[], int n, int k) { int lo = getMax(A, n); int hi = getSum(A, n); while (lo < hi) { int mid = lo + (hi-lo)/2; int requiredPainters = getRequiredPainters(A, n, mid); if (requiredPainters <= k) hi = mid; else lo = mid+1; } return lo; } class Solution { public: int shipWithinDays(vector<int>& weights, int D) { int num[50100] = {0}; int cnt = 0; int len = weights.size(); for(int i = 0 ; i < len ; i++){ num[i] = weights[i]; } return BinarySearch(num ,len, D); } };