class Solution {public double[] sampleStats(int[] count) {int len = 0;double[] res = new double[5];double sum = 0;double minNum = Integer.MAX_VALUE;double maxNum = Integer.MIN_VALUE;double meanNum = 0.0;double modeNum = 0.0;int modeCount = 0;for(int i=0; i<256; i++) {len += count[i];if( count[i]!=0 ) {maxNum = i;sum += ( count[i] * i );if( count[i]>modeCount ) {modeCount = count[i];modeNum = i;}}if( count[i]!=0 && minNum==Integer.MAX_VALUE ) minNum = i;}int cnt = 0;boolean flag = false;for(int i=0; i<256; i++) {cnt += count[i];if( cnt >= len/2 && flag == false ) {meanNum += i;flag = true;if( len%2 == 1 ) break;}if( cnt >= len/2+1 ) {meanNum += i;meanNum /= 2;break;}}res[0] = minNum;res[1] = maxNum;res[2] = (double) sum/len;res[3] = meanNum;res[4] = modeNum;return res;}
}

class Solution {public boolean carPooling(int[][] trips, int capacity) {int len = trips.length;int num = 0;Arrays.sort(trips, new Comparator<Object>() {  public int compare(Object oObjectA, Object oObjectB) {  int[] arTempOne = (int[])oObjectA;  int[] arTempTwo = (int[])oObjectB;   if (arTempOne[1] > arTempTwo[1]) {  return 1;  } else if (arTempOne[1] < arTempTwo[1]){  return -1;}return 0;  }});Map<Integer, Integer> map = new HashMap<>();int last = 0;for(int i=0; i<len; i++) {int num_passengers = trips[i][0];int start_location = trips[i][1];int end_location = trips[i][2];for(int j=last; j<=start_location; j++ ) {Integer tem = map.get(j);if( tem != null ) {num -= tem;}}last = start_location+1;Integer sum = map.get(end_location);if( sum == null ) {sum = num_passengers;} else sum += num_passengers;map.put(end_location, sum);num += num_passengers;if( num > capacity ) return false;}return true;}
}

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1095. Find in Mountain Array

You may recall that an array A is a mountain array if and only if:

• A.length >= 3
• There exists some i with 0 < i < A.length - 1 such that:
  • A[0] < A[1] < ... A[i-1] < A[i]
  • A[i] > A[i+1] > ... > A[A.length - 1]

Given a mountain array mountainArr, return the minimum index such that mountainArr.get(index) == target.  If such an index doesn't exist, return -1.

You can't access the mountain array directly.  You may only access the array using a MountainArray interface:

• MountainArray.get(k) returns the element of the array at index k (0-indexed).
• MountainArray.length() returns the length of the array.

Submissions making more than 100 calls to MountainArray.get will be judged Wrong Answer.  Also, any solutions that attempt to circumvent the judge will result in disqualification.

Example 1:

Input: array = [1,2,3,4,5,3,1], target = 3
Output: 2
Explanation: 3 exists in the array, at index=2 and index=5. Return the minimum index, which is 2.

Example 2:

Input: array = [0,1,2,4,2,1], target = 3
Output: -1
Explanation: 3 does not exist in the array, so we return -1.

Constraints:

1. 3 <= mountain_arr.length() <= 10000
2. 0 <= target <= 10^9
3. 0 <= mountain_arr.get(index) <= 10^9

/*** // This is MountainArray's API interface.* // You should not implement it, or speculate about its implementation* interface MountainArray {*     public int get(int index) {}*     public int length() {}* }*/class Solution {public int findInMountainArray(int target, MountainArray mountainArr) {int n = mountainArr.length();int top = -1;// 找顶点
            {int low = 0, high = n;while(high - low > 1){int h = high+low-1>>1;if(mountainArr.get(h) < mountainArr.get(h+1)){low = h+1;}else{high = h+1;}}top = low;}if(mountainArr.get(top) == target)return top;// 找左边
            {int low = 0, high = top+1;while(high - low > 0){int h = high+low>>1;int v = mountainArr.get(h);if(v == target)return h;if(v < target){low = h+1;}else{high = h;}}}// 找右边
            {int low = top, high = n;while(high - low > 0){int h = high+low>>1;int v = mountainArr.get(h);if(v == target)return h;if(v > target){low = h+1;}else{high = h;}}}return -1;}}