Given a sorted integer array where the range of elements are [lower, upper] inclusive, return its missing ranges.

For example, given [0, 1, 3, 50, 75], lower = 0 and upper = 99, return ["2", "4->49", "51->74", "76->99"].

[分析]

O(N) 扫一遍即可, 跟新LOWER,


[注意]

两层循环, 共用一个i时, 内层循环要注意判断边界条件.

[CODE]

public class Solution {
    public List<String> findMissingRanges(int[] A, int lower, int upper) {
        if(A==null) return null;
        List<String> res = new ArrayList<String>();
        for(int i=0; i<A.length; i++) {
            while(i<A.length && A[i] == lower ) {lower++; i++;} 
            if(i>=A.length) break;
            if(A[i] == lower+1) {
                res.add(String.valueOf(lower));
            } else {
                res.add("" + lower + "->" + (A[i]-1) );
            }
            lower = A[i] + 1;
        }
        
        if(lower == upper) {
            res.add(String.valueOf(lower));
        } else if(lower < upper ){
            res.add("" + lower + "->" + upper );
        }
        return res;
    }
}

leetcode 163: Missing Ranges