Given a sorted integer array without duplicates, return the summary of its ranges.
For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].
1
先看代码:
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public List<String> summaryRanges(int[] nums) {
List<String> list = new ArrayList<String>();
if(nums == null || nums.length == 0) return list;
int n = nums.length;
int start = 0;
int end = 0;
boolean isSearch = false;
for(int i = 0; i < n-1; i++){
if(isSearch == false) start = i;
if(nums[i+1] == nums[i] + 1){
isSearch = true;
}
else {
end = i;
isSearch = false;
String tmp = generate(nums,start,end);
list.add(tmp);
}
}
if(isSearch == false){
String tmp = generate(nums,n-1,n-1);
list.add(tmp);
}else{
String tmp = generate(nums,start,n-1);
list.add(tmp);
}
return list;
<span class="o">}</span>
<span class="kd">private</span> <span class="n">String</span> <span class="nf">generate</span><span class="o">(</span><span class="kt">int</span><span class="o">[]</span> <span class="n">nums</span><span class="o">,</span><span class="kt">int</span> <span class="n">left</span><span class="o">,</span><span class="kt">int</span> <span class="n">right</span><span class="o">){</span>
<span class="k">if</span><span class="o">(</span><span class="n">left</span> <span class="o">==</span> <span class="n">right</span><span class="o">)</span>
<span class="k">return</span> <span class="n">nums</span><span class="o">[</span><span class="n">left</span><span class="o">]+</span><span class="s">""</span><span class="o">;</span>
<span class="k">else</span><span class="o">{</span>
<span class="k">return</span> <span class="n">nums</span><span class="o">[</span><span class="n">left</span><span class="o">]+</span><span class="s">"->"</span><span class="o">+</span><span class="n">nums</span><span class="o">[</span><span class="n">right</span><span class="o">];</span>
<span class="o">}</span>
<span class="o">}</span><span class="w">
我是这么想的:使用start end两个变量标识连续元素的起点终点，使用isSearch表示当前位置是起点还是在连续元素之间。 初始化start=end=0,遍历元素，当isSearch==false时，说明开始了寻找新的一段连续元素，start置为当前的i。如果i+1的元素和i的元素相差1，则isSearch置为true,表示当前处于连续元素之间，当i+1和i的元素差值不为1，则将end置为i,isSearch置为false,表示这一段连续元素的起点是start,终点是end。注意一下最后的元素，当isSearch为false时，表示最后一个元素单独成为一段，当isSearch为true时，表示最后一个元素是以start为首的连续字符的最后一个元素。 (我发现我说半天还是说不清楚一个问题，想说的详细点却说的很啰嗦，争取多写，多表达自己的想法)
2 most votes
这个基本上就是这个思路。