算法学习day30-贪心算法

用最少数量的箭引爆气球

核心思路

这道题贪心的点在于, 让气球尽可能多的有重合的部分, 这样就可以一箭射穿
先根据左边界排序, 再判断下个气球的左边界和上个气球的右边界是否有重合的部分, 有的话, 就能一箭射穿
- 注意, 这里要更新重合部分的边界, 重合部分的边界是: $当前气球的左边界, 两个气球的最小的右边界$

如果没有重合的部分, 那就从当前气球开始重新寻找重合的边界

java 复制代码

public static int findMinArrowShots(int[][] points) {
    Arrays.sort(points, Comparator.comparingInt(a -> a[0]));
    int res = 1;
    int rightBian = points[0][1];
    for (int i = 1; i < points.length; i++) {
        int[] point = points[i];
        int left = point[0];
        int right = point[1];
        if (left > rightBian) {
            res++;
            rightBian = right;
        } else {
            rightBian = Math.min(right, rightBian);
        }
    }
    return res;
}

无重叠区间

核心思路

和射气球思路类似, 如果重叠就移除, 不重叠, 就用新的右边界继续判断是否重叠

java 复制代码

public int eraseOverlapIntervals(int[][] intervals) {
    Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
    int preRight = intervals[0][1];
    int count = 0;
    for (int i = 1; i < intervals.length; i++) {
        int[] ints = intervals[i];
        int left = ints[0];
        int right = ints[1];
        if (left < preRight) {
            count++;
            preRight = Math.min(right, preRight);
        } else {
            preRight = right;
        }
    }
    return count;
}

划分字母区间

核心思路

找到每个字母的最大区间范围, 如果下个字母和上个字母有重叠, 那就取最小left和最大right作为新的区间, 继续判断

java 复制代码

public List<Integer> partitionLabels(String s) {
    int left = 0;
    int right = 0;
    int[] fastIndex = new int[26];
    char[] charArray = s.toCharArray();
    for (int i = 0; i < charArray.length; i++) {
        fastIndex[charArray[i] - 'a'] = i;
    }
    List<Integer> res = new ArrayList<>();
    for (int i = 0; i < charArray.length; i++) {
        int index = fastIndex[charArray[i] - 'a'];
        if (i <= right) {
            left = Math.min(i, left);
            right = Math.max(index, right);
        } else {
            res.add(right - left + 1);
            left = i;
            right = index;
        }
    }
    res.add(right - left + 1);
    return res;
}