力扣labuladong一刷day18天差分数组

力扣labuladong一刷day18天差分数组

一、370. 区间加法

题目链接：https://leetcode.cn/problems/range-addition/
思路：这种频繁改变数组的值，最后在问数组中的值都是多少的情况特别适合使用查分数组，所谓查分数组，就是数组中每一个位置的值都等于原数组相邻位置相减的值，即diff[0]=nums[0]，diff[i] = nums[i]-nums[i-1]。
如nums = [a, b, c, d] diff = [a, b-a, c-b, d-c]
这样如果对区间[i, j] 加val，只需要对差分数组diff进行 diff[i]+=val ，diff[j+1]-=val。即可，这样不管做多少次的区间加值，只需要对区间首尾进行操作，时间复杂度就降低了。
那如何从差分数组还原回正常数组呢，只需要，res[0]=diff[0]，res[i] = res[i-1]+diff[i]。res定义为正常数组，上面的地推公式就相当于把每个位置减掉的前一个位置的值又加了回来。

int[] getModifiedArray(int length, int[][] updates) {
        int[] nums = new int[length];
        Difference difference = new Difference(nums);
        for (int[] update : updates) {
            difference.increment(update[0], update[1], update[2]);
        }
        return difference.result();
    }
class Difference {
    int[] diff;

    // 构造查分数组
    public Difference(int[] nums) {
        diff = new int[nums.length];
        diff[0] = nums[0];
        for (int i = 1; i < diff.length; i++) {
            diff[i] = nums[i] - nums[i-1];
        }
    }

    void increment(int i, int j, int val) {
        diff[i] += val;
        if (j+1 < diff.length) {
            diff[j+1] -= val;
        }
    }

    int[] result() {
        int[] nums = new int[diff.length];
        nums[0] = diff[0];
        for (int i = 1; i < diff.length; i++) {
            nums[i] = nums[i-1]+diff[i];
        }
        return nums;
    }
}

二、1109. 航班预订统计

题目链接：https://leetcode.cn/problems/corporate-flight-bookings/
思路：思路和上一题一样就是差分数组。

class Solution {
    public int[] corpFlightBookings(int[][] bookings, int n) {
        int[] nums = new int[n];
        Difference difference = new Difference(nums);
        for (int[] book : bookings) {
            difference.sum(book[0]-1, book[1]-1, book[2]);
        }
        return difference.result();
    }
}

class Difference {

    int[] diff;
    public Difference(int[] nums) {
        diff = new int[nums.length];
        diff[0] = nums[0];
        for (int i = 1; i < nums.length; i++) {
            diff[i] = nums[i] - nums[i-1];
        }
    }

    void sum(int i, int j, int val) {
        diff[i] += val;
        if (j+1 < diff.length) {
            diff[j+1] -= val;
        }
    }

    int[] result() {
        int[] res = new int[diff.length];
        res[0] = diff[0];
        for (int i = 1; i < diff.length; i++) {
            res[i] = res[i-1] + diff[i];
        }
        return res;
    }
}

三、1094. 拼车

题目链接：https://leetcode.cn/problems/car-pooling/
思路：这个拼车也是典型的差分数组的应用场景，区间[i, j]表示从i上a个人到 j 下来。其实不用考虑上下人的问题，就是把全部的区间都叠加到一起，叠加的部分都是在车上的，只要有一个地方大于容量即false。

class Solution {
    public boolean carPooling(int[][] trips, int capacity) {
        Difference difference = new Difference();
        for (int[] trip : trips) {
            difference.sum(trip[1], trip[2], trip[0]);
        }
        int[] result = difference.result();
        for (int i : result) {
            if (i > capacity) return false;
        }
        return true;
    }
}

class Difference {

    int[] diff = new int[1001];
    void sum(int i, int j, int val) {
        diff[i] += val;
        if (j + 1 < diff.length) {
            diff[j] -= val;
        }
    }
    int[] result() {
        int[] res = new int[diff.length];
        res[0] = diff[0];
        for (int i = 1; i < res.length; i++) {
            res[i] = res[i-1] + diff[i];
        }
        return res;
    }
}