保证每次操作都是局部最优的，并且最后得到的结果是全局最优的。

1. 分配饼干

455. Assign Cookies (Easy)

class Solution {
    public int findContentChildren(int[] g, int[] s) {
        if(g.length == 0 || s.length ==0){
            return 0;
        }
        Arrays.sort(g);
        Arrays.sort(s);
        int num = 0;
        int i=0;
        int j=0;
        while(i<g.length && j < s.length){
            if(g[i] <= s[j]){
                num++;
                i++;
                j++;
            }else{
                j++;
            }
        }
        return num;
    }
}

2. 不重叠的区间个数

435. Non-overlapping Intervals (Medium)

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if(intervals.length == 0){
            return 0;
        }
        Arrays.sort(intervals,Comparator.comparingInt(o -> o[1]));
        int cnt = 1;
        int end = intervals[0][1];
        for(int i=1; i < intervals.length; i++){
            if(intervals[i][0] < end){
                continue;
            }
            end = intervals[i][1];
            cnt++;
        }
        return intervals.length - cnt;
    }
}

3. 投飞镖刺破气球

452. Minimum Number of Arrows to Burst Balloons (Medium)

题目描述：气球在一个水平数轴上摆放，可以重叠，飞镖垂直投向坐标轴，使得路径上的气球都被刺破。求解最小的投飞镖次数使所有气球都被刺破。

也是计算不重叠的区间个数，不过和 Non-overlapping Intervals 的区别在于，[1, 2] 和 [2, 3] 在本题中算是重叠区间。

class Solution {
    public int findMinArrowShots(int[][] points) {
    if (points.length == 0) {
        return 0;
    }
    Arrays.sort(points, Comparator.comparingInt(o -> o[1]));
    int cnt = 1, end = points[0][1];
    for (int i = 1; i < points.length; i++) {
        if (points[i][0] <= end) {
            continue;
        }
        cnt++;
        end = points[i][1];
    }
    return cnt;
    }
}