【每日一题】Leetcode - 面试题 17.08. Circus Tower LCCI-Toy模板网

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Question

Leetcode - 面试题 17.08. Circus Tower LCCI

Train of thought

Sorting heights to be ascending order and weights to be descending order. dp[i] = j represents person[i] as the bottom of tower, the tower height is amount of j, to calculate the dp[i] we find the maximum of dp[0 ~ (i-1)] what the person[0 ~ (i-1)] shorter and lighter than person[i], increase it of one, it’s the dp[i] result.

class Solution {

    public int bestSeqAtIndex(int[] height, int[] weight) {
        int[][] persons = new int[height.length][2];
        for (int i = 0; i < height.length; i++) {
            persons[i][0] = height[i];
            persons[i][1] = weight[i];
        }
        Arrays.sort(persons, (a, b) -> {
            if (a[0] == b[0]) {
                return b[1] - a[1];
            }
            return a[0] - b[0];
        });
        int[] dp = new int[persons.length];
        Arrays.fill(dp, 1);
        int max = 1;
        for (int i = 1; i < persons.length; i++) {
            for (int j = i - 1; j >= 0; j--) {
                if (persons[j][1] < persons[i][1]) {
                    if (dp[j] + 1 > dp[i]) {
                        dp[i] = dp[j] + 1;
                    }
                }
            }
            if (dp[i] > max) {
                max = dp[i];
            }
        }
        return max;
    }

}

【每日一题】Leetcode - 面试题 17.08. Circus Tower LCCI,每日一题,leetcode,算法,二分搜索,java,动态规划

Optimize

We can optimize the follow code, make it time complexity from O(n) to O(logn):

	for (int j = i - 1; j >= 0; j--) {
       if (persons[j][1] < persons[i][1]) {
            if (dp[j] + 1 > dp[i]) {
                dp[i] = dp[j] + 1;
            }
        }
    }
    if (dp[i] > max) {
        max = dp[i];
    }
    
//-------------------- TO BE ------------------

	 left = 0; right = len;
     while (left < right) {
         mid = (left + right) / 2;
         if (top[mid] > persons[i][1]) {
             right = mid;
         }
         else if (top[mid] < persons[i][1]) {
             left = mid + 1;
         }
         else {
             right = mid;
         }
     }
     if (left == len) {
         len++;
     }
     top[left] = persons[i][1];

Following is the complete code

class Solution {

    public int bestSeqAtIndex(int[] height, int[] weight) {
        int[][] persons = new int[height.length][2];
        for (int i = 0; i < height.length; i++) {
            persons[i][0] = height[i];
            persons[i][1] = weight[i];
        }
        Arrays.sort(persons, (a, b) -> {
            if (a[0] == b[0]) {
                //  using descending order to avoid choice twice person of the weights are equals,
                //      the principle is because of we choice the continuous top is ascending order
                //      and the heap is descending order, it can covered the ahead
                return b[1] - a[1];
            }
            return a[0] - b[0];
        });
        int[] top = new int[persons.length];
        int len = 0, left, right, mid;
        for (int i = 0; i < persons.length; i++) {
            left = 0; right = len;
            while (left < right) {
                mid = (left + right) / 2;
                if (top[mid] > persons[i][1]) {
                    right = mid;
                }
                else if (top[mid] < persons[i][1]) {
                    left = mid + 1;
                }
                else {
                    right = mid;
                }
            }
            if (left == len) {
                len++;
            }
            top[left] = persons[i][1];
        }
        return len;
    }

}